GIFT  OF. 


SOLID   GEOMETRY 


BY 
C.   A.   HART 

INSTRUCTOR   OF   MATHEMATICS,   WADLEI^H    HIGH   SCHOOL,   NEW   YORK   CITY 

AND 

DANIEL   D.   FELDMAN 

HEAD   OF  DEPARTMENT  OF   MATHEMATICS,    ERASMUS   HALL  HIGH   SCHOOL,   BROOKLYN 

WITH  THE   EDITORIAL  COOPERATION  OF 
J.   H.   TANNER  AND   VIRGIL   SNYDER 

PROFESSORS   OF   MATHEMATICS   IN   CORNELL   UNIVERSITY 


NEW  YORK  •:•  CINCINNATI  •:•  CHICAGO 

AMERICAN    BOOK    COMPANY 


\[ 


COPYRIGHT,  1912,  BY 
AMERICAN  BOOK  COMPANY. 

ENTERED  AT  STATIONERS'  HALL,  LONDON. 


H.-F.   SOLID  GEOMETRY. 
W.  P.     I 


PREFACE 

IN  addition  to  the  features  of  the  Plane  Geometry,  which  are 
emphasized  in  the  Solid  as  well,  the  chief  characteristic  of  this 
book  is  the  establishment,  at  every  point,  of  the  vital  relation 
between  the  Solid  and  the  Plane  Geometry.  Many  theorems  in 
Solid  Geometry  have  been  proved,  and  many  problems  have 
been  solved,  by  reducing  them  to  a  plane,  and  simply  applying 
the  corresponding  principle  of  Plane  Geometry.  Again,  many 
proofs  of  Plane  Geometry  have  been  made  to  serve  as  proofs 
of  corresponding  theorems  in  Solid  Geometry  by  merely  mak- 
ing the  proper  changes  in  terms  used.  (See  §§  703,  786,  794, 
813,  853,  924,  951,  955,  961,  etc.) 

Other  special  features  of  the  book  may  be  summarized  as 
follows  : 

The  student  is  given  every  possible  aid  in  forming  his  early 
space  concepts.  In  the  early  work  in  Solid  Geometry,  the 
average  student  experiences  difficulty  in  fully  comprehending 
space  relations ;  that  is,  in  seeing  geometric  figures  in  space. 
The  student  is  aided  in  o/e.Toroing  this  difficulty  by  the  intro- 
duction of  many  easy  and1  practical  questi6risj  and  exercises,  as 
well  as  by  being  encouraged  to  make  his  frgirrae.  (See  §  605.)  As 
a  further  aid  in  this  direction,  reproductions  of  models  made 
by  students  themselves  are  shown  in  a  group  (p.  302),  and  at 
various  points  throughout  Book  VI. 

The  student's  fund  of  knowledge  is  constantly  drawn  upon.  In 
the  many  questions,  suggestions,  and  exercises,  his  knowledge 
of  the  things  about  him  has  been  constantly  appealed  to. 
Especially  is  this  true  pf  the  work  on  the  sphere,  where  the 

sa 
249124 


iv  PREFACE 

student's  knowledge  of  mathematical  geography  has  been  ap- 
pealed to  in  making  clear  the  terms  and  the  relations  of  figures 
connected  with  the  sphere. 

The  treatment  of  the  Solid  Geometry  is  logical.  The  same 
logical  rigor  that  characterizes  the  demonstrations  in  the  Plane 
Geometry  is  used  consistently  throughout  the  Solid.  If  a  pos- 
tulate is  needed  to  make  a  proof  complete,  it  is  clearly  stated, 
as  in  §  615.  In  the  mensuration  of  the  prism  and  the  pyra- 
mid, the  same  general  plan  has  been  followed  as  that  used  in 
Book  IV;  in  the  mensuration  of  the  cylinder,  the  cone,  and 
the  sphere,  the  method  pursued  is  similar  to  that  used  in  the 
mensuration  of  the  circle.  ' 

More  proofs  and  parts  of  proofs  are  left  to  the  student  in  the 
Solid,  than  in  the  Plane  Geometry ;  but  in  every  case  in  which 
the  proof  is  not  complete,  the  incompleteness  is  specifically 
stated. 

TJie  treatment  of  the  polyhedral  angle  (p.  336),  of  the  prism 
(p.  345),  and  of  the  pyramid  (p.  350),  is  similar  to  that  of  the 
cylinder  and  the  cone.  This  is  in  accordance  with  the  recom- 
mendations of  the  leading  Mathematical  Associations  through- 
out the  country. 

The  complete  collection  of  formulas  of  Solid  Geometry  at  the 
end  of  the  book,  it  is  hoped,  will  be  found  helpful  to  teacher 
and  student  alike. 

The  grateful  acknowledgment  of  the  authors  is  due  to  many 
friends  for  helpful '  suggestions  ;  especially  to  Miss  Grace  A. 
Brace,  of  the  Wadleigk  'High  School,  New  York ;  to  Mr. 
Edward  B.  PArsJofus:  of  th-3  .Boys'  High  School,  Brooklyn ;  and 
to  Professor'Mc'Mahbri,  of  Cornell  University. 


CONTENTS 

SOLID   GEOMETRY 

PAGE 

SYMBOLS  AND   ABBREVIATIONS     .         .         .         .         .  vi 

REFERENCES   TO   PLANE   GEOMETRY vii 

BOOK  VI.     LINES,   PLANES,    AND   ANGLES   IN  SPACE      .  299 

Lines  and  Planes         .........  301 

Dihedral  Angles 322 

Polyhedral  Angles 336 

BOOK   VII.     POLYHEDRONS 343 

Prisms 345 

Pyramids 350 

Mensuration  of  the  Prism  and  Pyramid 354 

Areas .354 

Volumes  .  .  t 358 

Miscellaneous  Exercises       ........  381 

BOOK  VIII.     CYLINDERS  AND   CONES 383 

Cylinders 383 

Cones  .                          388 

Mensuration  of  the  Cylinder  and  Cone        .....  392 

Areas 392 

Volumes 406 

Miscellaneous  Exercises 414 

BOOK   IX.     THE    SPHERE 417 

Lines  and  Planes  Tangent  to  a  Sphere 424 

Spherical  Polygons .        .         .429 

Mensuration  of  the  Sphere  ........  444 

Areas    ...........  444 

Volumes 457 

Miscellaneous  Exercises  on  Solid  Geometry        ....  467 

FORMULAS   OF   SOLID    GEOMETRY 471 

APPENDIX  TO   SOLID   GEOMETRY 474 

Spherical  Segments     .........  474 

The  Prismatoid   .         .        ,        .......  475 

Similar  Polyhedrons    .         .         .        .                 .         .         .         .  477 

INDEX 481 

v 


SYMBOLS   AND   ABBREVIATIONS 

=  equals,  equal  to,  is  equal  to.  rt.  right. 

=£  does  not  equal.  str.  straight. 

>  greater  than,  is  greater  than.  ext.  exterior. 

<  less  than,  is  less  than.  int.  interior. 

~  equivalent,   equivalent  to,  is  equiva-      alt.  alternate, 

lent  to.  def.  definition. 

~  similar,  similar  to,  is  similar  to.  ax.  axiom. 

S2  is  measured  by.  post.  postulate. 

_L  perpendicular,    perpendicular    to,    is      hyp.  hypothesis, 

perpendicular  to.  prop,  proposition. 

Js  perpendiculars.  prob.  problem. 

||  parallel,  parallel  to,  is  parallel  to.  th.  theorem. 

||s  parallels.  cor.  corollary. 

.  .  .  and  so  on  (sign  of  continuation).  cons,  construction. 

•••  since.  ex.  exercise. 

.•.  therefore.  fig.  figure. 

~  arc  ;  AB,  arc  AB.  iden.  identity. 

O,  [U  parallelogram,  parallelograms.  comP-  complementary. 

O,  ©  circle,  circles.  sup.  supplementary. 

Z,  A  angle,  angles.  adJ-  adjacent. 

A,  &  triangle,  triangles.  homol.  homologous. 

Q.E.D.     Quod  erat  demonstrandum,  which  was  to  be  proved. 

Q.E.F.     Quod  erat  faciendum,  which  was  to  be  done. 

The  signs  -fj  — ,  x  ,  -f-  have  the  same  meanings  as  in  algebra. 


vi 


REFERENCES  TO  THE  PLANE  GEOMETRY 

Note.  The  following  definitions,  theorems,  etc.,  from  the  Plane 
Geometry  which  are  referred  to  in  the  Solid  Geometry  are  here  collected 
for  the  convenience  of  the  student. 

(The  numbers  below  refer  to  articles  in  the  Plane  Geometry.) 

18.  Def.  Two  geometric  figures  are  equal  if  they  can  be  made  to 
coincide. 

26.  Two  intersecting  straight  lines  can  have  only  one  point  in 
common;  i.e.  two  intersecting  straight  lines  determine  a  point. 

34.  Def.  A  plane  surface  (or  plane)  is  a  surface  of  unlimited 
extent  such  that  whatever  two  of  its  points  are  taken,  a  straight  line 
joining  them  will  lie  wholly  in  the  surface. 

ASSUMPTIONS 

54.  1.  Things  equal  to  the  same  thing,  or  to  equal  things,  are 
equal  to  each  other. 

2.  If  equals  are  added  to  equals,  the  sums  are  equal. 

3.  If  equals  are  subtracted  from  equals,  the  remainders  are  equal. 

4.  If  equals  are  added  to  unequal s,  the  sums  are  unequal  in  the 
same  order. 

5.  If  equals   are  subtracted  from  unequals,  the  remainders  are 
unequal  in  the  same  order. 

6.  If  unequals  are  subtracted  from  equals,  the  remainders  are 
unequal  in  the  reverse  order. 

7.  (a)    If  equals  are  multiplied  by  equals,  the  products  are  equal ; 

(5)  if  unequals  are  multiplied  by  equals,  the  products  are  unequal  in 
the  same  order. 

8.  (a)    If  equals  are  divided  by  equals,  the  quotients  are  equal ; 

(6)  if  unequals  are  divided  by  equals,  the  quotients  are  unequal  in 
the  same  order. 

9.  If  unequals  are  added  to  unequals,  the  less  to  the  less  and  the 
greater  to  the  greater,  the  sums  are  unequal  in  the  same  order. 

10.  If  three  magnitudes  of  the  same  kind  are  so  related  that  the 
first  is  greater  than  the  second,  and  the  second  greater  than  the  third, 
then  the  first  is  greater  than  the  third. 

vii 


viii        REFERENCES   TO   THE   PLANE   GEOMETRY 

11.  The  whole  is  equal  to  the  sum  of  all  its  parts. 

12.  The  whole  is  greater  than  any  of  its  parts. 

13.  Like   powers   of  equal  numbers  are  equal,  and   like  roots  of 
equal  numbers  are  equal. 

14.  Transference    postulate.     Any    geometric    figure    may    be 
moved  from    one   position   to   another   without  change    of    size    or 
shape. 

15.  Straight  line  postulate  I.     A  straight  line  may  be   drawn 
from  any  one  point  to  any  other. 

16.  Straight  line  postulate  II.     A   line  segment  may  be   pro- 
longed indefinitely  at  either  end. 

17.  Revolution   postulate.     A  straight  line  may  revolve   in   a 
plane,  about  a  point  as  a  pivot,  and  when  it  does  revolve  continuously 
from  one  position  to  another,  it  passes  once  and  only  once  through 
every  intermediate  position.. 

62.  At  every  point  in  a  straight  line  there  exists  only  one  perpen- 
dicular to  the  line. 

63.  At  every  point  in  a  straight  line  there  exists  one  and  only  one 
perpendicular  to  the  line. 

65.  If  one  straight  line  meets  another  straight  line,  the  sum  of  the 
two  adjacent  angles  is  two  right  angles. 

76.  If  two  adjacent  angles  are  supplementary,  their  exterior  sides 
are  collinear. 

77.  If  two  straight  lines  intersect,  the  vertical  angles  are  equal. 
92.   Def.     A  polygon  of   three  sides  is  called  a  triangle;  one  of 

four  sides,  a  quadrilateral;  one  of  five  sides,  a  pentagon;  one  of 
six  sides,  a  hexagon;  and  so  on. 

105.  Two  triangles  are  equal  if  a  side  and  the  two  adjacent 
angles  of  one  are  equal  respectively  to  a  side  and  the  two  adjacent 
angles  of  the  other. 

107.  Two  triangles  are  equal  if  two  sides  and  the  included  angle 
of  one  are  equal  respectively  to  two  sides  and  the  included  angle  of 
the  other. 

110.  Homologous  parts  of  equal  figures  are  equal. 

111.  The  base  angles  of  an  isosceles  triangle  are  equal. 

116.  Two  triangles  are  equal  if  the  three  sides  of  one  are  equal 
respectively  to  the  three  sides  of  the  other, 


REFERENCES  TO   THE  PLANE   GEOMETRY  ix 

122.  Circle  postulate.  A  circle  may  be  constructed  having  any 
point  as  center,  and  having  a  radius  equal  to  any  finite  line. 

124.  To  construct  an  equilateral  triangle,  with  a  given  line  as 
side. 

134.  Every  point  in  the  perpendicular  bisector  of  a  line  is  equi- 
distant from  the  ends  of  that  line. 

139.  Every  point  equidistant  from  the  ends  of  a  line  lies  in  the 
perpendicular  bisector  of  that  line. 

142.  Two  points  each  equidistant  from  the  ends  of  a  line  deter- 
mine the  perpendicular  bisector  of  the  line. 

148.  To  construct  a  perpendicular  to  a  given  straight  line  at  a 
given  point  in  the  line. 

149.  From  a  point  outside  a  line  to  construct  a  perpendicular  to 
the  line. 

153.  If  one  side  of  a  triangle  is  prolonged,  the  exterior  angle 
formed  is  greater  than  either  of  the  remote  interior  angles. 

154.  From  a  point  outside  a  line  there  exists  only  one  perpendic- 
ular to  the  line. 

156.  If  two  sides  of  a  triangle  are  unequal,  the  angle  opposite  the 
greater  side  is  greater  than  the  angle  opposite  the  less  side. 

161.  («)  In  the  use  of  the  indirect  method  the  student  should  give, 
as  argument  1,  all  the  suppositions  of  which  the  case  he  is  considering 
admits,  including  the  conclusion.  As  reason  1  the  number  of  such 
possible  suppositions  should  be  cited. 

(&)  As  a  reason  for  the  last  step  in  the  argument  he  should  state 
which  of  these  suppositions  have  been  proved  false. 

167.  The  sum  of  any  two  sides  of  a  triangle  is  greater  than  the 
third  side. 

168.  Any  side  of  a  triangle  is  less  than  the  sum  and  greater  than 
the  difference  of  the  other  two. 

173.  If  two  triangles  have  two  sides  of  one  equal  respectively  to 
two  sides  of  the  other,  but  the  third  side  of  the  first  greater  than  the 
third  side  of  the  second,  then  the  angle  opposite  the  third  side  of  the 
first  is  greater  than  the  angle  opposite  the  third  side  of  the  second. 

178.  Parallel  line  postulate.  Two  intersecting  straight  lines 
cannot  both  be  parallel  to  the  same  straight  line. 


X  REFERENCES   TO  THE   PLANE   GEOMETRY 

179.  The  following  form  of  this  postulate  is  sometimes  more  con- 
venient to  quote :  Through  a  given  point  there  exists  only  one  line 
parallel  to  a  given  line. 

180.  If  two  straight  lines  are  parallel  to  a  third  straight  line,  they 
are  parallel  to  each  other. 

187.  If  two  straight  lines  are  perpendicular  to  a  third  straight  line, 
they  are  parallel  to  each  other. 

190.  If  two  parallel  lines  are  cut  by  a  transversal,  the  correspond- 
ing angles  are  equal. 

192.  If  two  parallel  lines  are  cut  by  a  transversal,  the  sum  of  the 
two  interior  angles  on  the  same  side  of  the  transversal  is  two  right 
angles. 

193.  A  straight  line  perpendicular  to  one  of  two  parallels  is  per- 
pendicular to  the  other  also. 

194.  If  two  straight  lines    are  cut  by  a  transversal  making   the 
sum  of  the  two  interior  angles  on  the  same  side  of  the  transversal  not 
equal  to  two  right  angles,  the  lines  are  not  parallel. 

198.  Two  angles  whose  sides  are  parallel,  each  to  each,  are  either 
equal  or  supplementary. 

206.  In  a  triangle  there  can  be  but  one  right  angle  or  one  obtuse 
angle. 

209.  Two  right  triangles  are  equal  if  the  hypotenuse  and  an 
acute  angle  of  one  are  equal  respectively  to  the  hypotenuse  and  an 
acute  angle  of  the  other. 

211.  Two  right  triangles  are  equal  if  the  hypotenuse  and  a  side 
of  one  -are  equal  respectively  to  the  hypotenuse  and  a  side  of  the 
other. 

215.  An  exterior  angle  of  a  triangle  is  equal  to  the  sum  of  the 
two  remote  interior  angles. 

^216.     The  sum  of  all  the  angles  of  any  polygon  is  twice  as  many 
right  angles  as  the  polygon  has  sides,  less  four  right  angles. 

220.  Def.  A  parallelogram  is  a  quadrilateral  whose  opposite 
sides  are  parallel. 

228.  Def.  Any  side  of  a  parallelogram  may  be  regarded  as  its 
base,  and  the  line  drawn  perpendicular  to  the  base  from  any  point 
in  the  opposite  side  is  then  the  altitude. 

232.     The  opposite  sides  of  a  parallelogram  are  equal. 


REFERENCES  TO   THE   PLANE   GEOMETRY  xi 

234.  Parallel  lines  intercepted  between  the  same  parallel  lines 
are  equal. 

240.  If  two  opposite  sides  of  a  quadrilateral  are  equal  and  par- 
allel, the  figure  is  a  parallelogram. 

252.  The  two  perpendiculars  to  the  sides  of  an  angle  from  any 
point  in  its  bisector  are  equal. 

253.  Every  point  in  the  bisector  of  an  angle  is  equidistant  from 
the  sides  of  the  angle. 

258.  The  bisectors  of  the  angles  of  a  triangle  are  concurrent  in  a 
point  which  is  equidistant  from  the  three  sides  of  the  triangle. 

276.  Def.  A  circle  is  a  plane  closed  figure  whose  boundary  is 
a  curve  such  that  all  straight  lines  to  it  from  a  fixed  point  within  are 
equal. 

279.     (a)    All  radii  of  the  same  circle  are  equal. 
(&)    All  radii  of  equal  circles  are  equal, 
(c)    All  circles  having  equal  radii  are  equal. 

297.  Four  right  angles  contain  360,  angle  degrees,  and  four  right 
angles  at  the  center  of  a  circle  intercept  a  complete  circumference ; 
therefore,  a  circumference  contains  360  arc  degrees.     Hence,  a  semi- 
circumference  contains  180  arc  degrees. 

298.  In  equal  circles,  or  in  the  same  circle,  if  two  chords  are  equal, 
they  subtend  equal  arcs ;  conversely,  if  two  arcs  are  equal,  the  chords 
that  subtend  them  are  equal. 

307.  In  equal  circles,  or  in  the  same  circle,  if  two  chords  are  equal, 
they  are  equally  distant  from  the  center ;  conversely,  if  two  chords  are 
equally  distant  from  the  center,  they  are' equal. 

308.  In  equal  circles,  or  in  the  same  circle,  if  two  chords  are 
unequal,  the  greater  chord  is  at  the  less  distance  from  the  center. 

309.  Note.     The  student  should  always  give  the  full   statement 
of  the  substitution  made ;    for  example,    "  Substituting  A  E  for  its 
equal  CD." 

310.  In  equal  circles,  or  in  the  same  circle,  if  two  chords  are  une- 
qually distant  from  the  center,  the  chord  at  the  less  distance  is  the 
greater. 

313.  A  tangent  to  a  circle  is  perpendicular  to  the  radius  drawn 
to  the  point  of  tangency. 


REFERENCES   TO   THE   PLANE   GEOMETRY 

314.  A  straight  line  perpendicular  to  a  radius  at  its  outer  ex- 
tremity is  tangent  to  the  circle. 

321.     To  inscribe  a  circle  in  a  given  triangle. 

323.  To  circumscribe  a  circle  about  a  given  triangle. 

324.  Three    points  not  in    the   same  straight   line  determine  a 
circle. 

328.  If  two  circumferences  intersect,  their  line  of  centers  bisects 
their  common  chord  at  right  angles. 

335.  Def.  To  measure  a  quantity  is  to  find  how  many  times  it 
contains  another  quantity  of  the  same  kind.  The  result  of  the 
measurement  is  a  number  and  is  called  the  numerical  measure,  or 
measure-number,  of  the  quantity  which  is  measured.  The  measure 
employed  is  called  the  unit  of  measure. 

337.  Def.  Two  quantities  are  commensurable  if  there  exists  a 
measure  that  is  contained  an  integral  number  of  times  in  each. 
Such  a  measure  is  called  a  common  measure  of  the  two  quantities. 

339.  Def.  Two  quantities  are  incommensurable  if  there  exists 
no  measure  that  is  contained  an  integral  number  of  times  in  each. 

341.  Def.  The  ratio  of  two  geometric  magnitudes  may  be 
defined  as  the  quotient  of  their  measure-numbers,  when  the  same 
measure  is  applied  to  each. 

349.  Def.  If  a  variable  approaches  a  constant  in  such  a  way  that 
the  difference  between  the  variable  and  the  constant  may  be  made  to 
become  and  remain  smaller  than  any  fixed  number  previously 
assigned,  however  small,  the  constant  is  called  the  limit  of  the 
variable. 

355.  If  two  variables  are  always  equal,  and  if  each  approaches  a 
limit,  then  their  limits  are  equal. 

358.  An  angle  at  the  center  of  a  circle  is  measured  by  its  inter- 
cepted arc. 

362.  (a)  In  equal  circles,  or  in  the  same  circle,  equal  angles  are 
measured  by  equal  arcs ;  conversely,  equal  arcs  measure  equal  angles. 

(I)     The  measure  of  the  -)  ,  J-  of  two  angles  is  equal  to 

J  difference  v 

the  }  ,  >  of  the  measures  of  the  angles. 

J  difference  ( 


REFERENCES   TO   THE   PLANE   GEOMETRY        xiii 

(c)  The  measure  of  any  multiple  of  an  angle  is  equal  to  that  same 
multiple  of  the  measure  of  the  angle. 

373.     To  construct  a  tangent  to  a  circle  from  a  point  outside. 

399.  If  four  numbers  are  in  proportion,  they  are  in  proportion  by 
division ;  that  is,  the  difference  of  the  first  two  terms  is  to  the  first 
(or  second)  term  as  the  difference  of  the  last  two  terms  is  to  the  third 
(or  fourth)  term. 

401.  In  a  series  of  equal  ratios  the  sum  of  any  number  of 
antecedents  is  to  the  sum  of  the  corresponding  consequents  as  any 
antecedent  is  to  its  consequent. 

409.  A  straight  line  parallel  to  one  side  of  a  triangle  divides  the 
other  two  sides  proportionally. 

410.  A  straight  line  parallel  to  one  side  of  a  triangle  divides  the 
other  two  sides  into  segments  which  are  proportional. 

419.  Def.     Two    polygons    are    similar    if    they    are    mutually 
equiangular  and  if  their  sides  are  proportional. 

420.  Two  triangles  which  are  mutually  equiangular  are  similar. 

422.  Two  right  triangles  are  similar  if  an  acute  angle  of  one  is 
equal  to  an  acute  angle  of  the  other. 

424.     (1)     Homologous  angles  of  similar  triangles  are  equal. 

(2)  Homologous  sides  of  similar  triangles  are  proportional. 

(3)  Homologous   sides  of  similar  triangles  are  the  sides 
opposite  equal  angles. 

435.  In  two  similar  triangles  any  two  homologous  altitudes  have 
the  same  ratio  as  any  two  homologous  sides. 

438.  If  two  polygons  are  composed  of  the  same  number  of  triangles, 
similar  each  to  each  and  similarly  placed,  the  polygons  are  similar. 

443.  In  a  right  triangle,  if  the  altitude  upon  the  hypotenuse  is 
drawn : 

I.  The  square  of  the  altitude  is  equal  to  the  product  of  the  seg- 
ments of  the  hypotenuse. 

II.  The  square  of  either  side  is  equal  to  the  product  of  the  whole 
hypotenuse  and  the  segment  of  the  hypotenuse  adjacent  to  that  side. 

444.  If  from  any  point  in  the  circumference  of  a  circle  a  per- 
pendicular to  a  diameter  is  drawn,  and  if  chords  are  drawn  from  the 
point  to  the  ends  of  the  diameter : 


xiv         REFERENCES   TO   THE   PLANE   GEOMETRY 

I.  The   perpendicular  is  a  mean  proportional   between   the   seg- 
ments of  the  diameter. 

II.  Either    chord   is    a    mean    proportional    between    the    whole 
diameter  and  the  segment  of  the  diameter  adjacent  to  the  chord. 

478.  The  area  of  a  square  is  equal  to  the  square  of  its  side. 

479.  Any  two  rectangles  are  to  each   other  as  the  products  of 
their  bases  and  their  altitudes. 

480.  (a)    Two  rectangles  having  equal  bases  are  to  each  other 
as  their  altitudes,  and  (b)  two  rectangles  having  equal  altitudes  are  to 
each  other  as  their  bases. 

481.  The  area  of  a  parallelogram  equals  the  product  of  its  base 
and  its  altitude. 

482.  Parallelograms   having  equal  bases  and  equal  altitudes  are 
equivalent. 

483.  Any  two  parallelograms  are  to  each  other  as  the  products  of 
their  bases  and  their  altitudes. 

484.  (a)    Two   parallelograms   having   equal   bases  are  to  each 
other  as  their  altitudes,  and  (b)  two  parallelograms   having  equal 
altitudes  are  to  each  other  as  their  bases. 

485.  The  area  of  a  triangle  equals  one  half  the  product  of  its  base 
and  its  altitude. 

491.  The  area  of  a  triangle  is  equal  to  one  half  the  product  of  its 
perimeter  and  the  radius  of  the  inscribed  circle. 

492.  The  area  of  any  polygon   circumscribed  about  a  circle  is 
equal  to  one  half  its  perimeter   multiplied  by  the  radius  of  the  in- 
scribed circle. 

498.  Two  triangles  which  have  an  angle  of  one  equal  to  an  angle 
of  the  other  are  to  each  other  as  the  products  of  the  sides  including 
the  equal  angles. 

503.  Two  similar  triangles  are  to  each  other  as  the  squares  of  any 
two  homologous  sides. 

517.  If  the  circumference  of  a  circle  is  divided  into  any  number  of 
equal  arcs  :  (a)  the  chords  joining  the  points  of  division  form  a  regu- 
lar polygon  inscribed  in  the  circle ;  (6)  tangents  drawn  at  the  points 
of  division  form  a  regular  polygon  circumscribed  about  the  circle. 


REFERENCES   TO   THE   PLANE   GEOMETRY         xv 

538.  The  perimeters  of  two  regular  polygons  of  the  same  number 
of  sides  are  to  each  other  as  their  radii  or  as  their  apothems. 

541.  I.  The  perimeter  and  area  of  a  regular  polygon  inscribed  in 
a  circle  are  less,  respectively,  than  the  perimeter  and  area  of  the  regu- 
lar inscribed  polygon  of  twice  as  many  sides. 

II.  The  perimeter  and  area  of  a  regular  polygon  circumscribed 
about  a  circle  are  greater,  respectively,  than  the  perimeter  and  area  of 
the  regular  circumscribed  polygon  of  twice  as  many  sides. 

543.  By  repeatedly  doubling  the  number  of  sides  of  a  regular  poly- 
gon inscribed  in  a  circle,  and  making  the  polygons  always  regular : 

I.  The  apothem  can  be  made  to  differ  from  the  radius  by  less  than 
any  assigned  value. 

II.  The  square  of  the  apothem  can  be  made  to  differ  from  the 
square  of  the  radius  by  less  than  any  assigned  value. 

546.  By  repeatedly  doubling  the  munber  of  sides  of  regular  cir- 
cumscribed and  inscribed  potygons  of  the  same  number  of  sides,  and 
making  the  polygons  always  regular  : 

I.  Their  perimeters  approach  a  common  limit. 

II.  Their  areas  approach  a  common  limit. 

550.  Def.  The  length  of  a  circumference  is  the  common  limit 
which  the  successive  perimeters  of  inscribed  and  circumscribed  regu- 
lar polygons  (of  3,  4,  5,  etc.,  sides)  approach  as  the  number  of  sides  is 
successively  increased  and  each  side  approaches  zero  as  a  limit. 

556.     Any  two  circumferences  are  to  each  other  as  their  radii. 

558.  Def.     The  area  of  a  circle  is  the  common  limit  which  the 
successive  areas  of  inscribed  and  circumscribed  regular  polygons  ap- 
proach as  the  number  of  sides  is  successively  increased  and  each  side 
approaches  zero  as  a  limit. 

559.  The  area  of  a  circle  is  equal  to  one  half  the  product  of  its 
circumference  and  its  radius. 

586.  If  a  variable  can  be  made  less  than  any  assigned  value,  the 
quotient  of  the  variable  by  any  constant,  except  zero,  can  be  made  less 
than  any  assigned  value. 

587.  If  a  variable  can  be  made  less  than  any  assigned  value,  the 
product  of  that  variable  and  a  decreasing  value  may  be  made  less 
than  any  assigned  value. 


xvi         REFERENCES   rl  O   THE   PLANE   GEOMETRY 

• 

590.     The  limit  of  the  product  of  a  variable  and  a  constant,  not 
zero,  is  the  limit  of  the  variable  multiplied  by  the  constant. 

592.  If  two  variables  approach  finite  limits,  not  zero,  then  the 
limit  of  their  product  is  equal  to  the  product  of  their  limits. 

593.  If   each   of    any  finite  number  of    variables   approaches   a 
finite  limit,  not  zero,  then  the  limit  of  their  product  is  equal  to  the 
product  of  their  limits. 

594.  If  two  related  variables  are  such  that  one  is  always  greater 
than  the  other,  and  if  the  greater  continually  decreases  while  the  less 
continually  increases,  so  that  the  difference  between  the  two  may  be 
made  as  small  as  we  please,  *then  the  two  variables  have  a  common 
limit  which  lies  between  them. 

599.     An  angle  can  be  bisected  by  only  one  line. 


SOLID   GEOMETRY 

BOOK  VI 

LINES,  PLANES,  AND  ANGLES  IN  SPACE 

602.  Def .     Solid  geometry  or  the  geometry  of  space  treats  of 
figures  whose  parts  are  not  all  in  the  same  plane.     (For  defini- 
tion of  plane  or  plane  surface,  see  §  34.) 

603.  From  the  definition  of  a  plane  it  follows  that : 

\ptylftwo  points  of  a  straight  line  lie  in  a  plane,  the  whole  line 
lies  in  that  plane. 

(b)  A  straight  line  can  intersect  a  plane  in  not  more  than  one 
point. 

604.  Since   a   plane   is   unlimited   in  its    two   dimensions 

(length    and    breadth)  __ 

only  a  portion  of  it  can 

be  shown  in  a  figure. 
This  is  usually  repre- 
sented    by     a     quad- 
rilateral  drawn    as    a    j 
parallelogram.       Thus 

MN  represents  a  plane.  Sometimes,  however,  conditions  make 
it  necessary  to  represent  a  plane  by  a  figure  other  than  a 
parallelogram,  as  in  §  617. 


Ex.  1152.  Draw  a  rectangle  freehand  which  is  supposed  to  lie  :  (a) 
in  a  vertical  plane  ;  (6)  in  a  horizontal  plane.  May  the  four  angles  of 
the  rectangle  of  (a)  be  drawn  equal  ?  those  of  the  rectangle  of  (6)  ? 

299 


300  '  '    '   SOLID  GEOMETRY 

605. '  Note.  In  tlfe'figMres  in  solid  geometry  dashed  lines  will  be  used 
to  represent  all  auxiliary  lines  and  lines  that  are  not  supposed  to  be  visible 
but  which,  for  purposes  of  proof,  are  represented  in  the  figure.  All  other 
lines  will  be  continuous.  In  the  earlier  work  in  solid  geometry  the  stu- 
dent may  experience  difficulty  in  imagining  the  figures.  If  so,  he  may 
find  it  a  great  help,  for  a  time  at  least,  to  make  the  figures.  By  using 
pasteboard  to  represent  planes,  thin  sticks  of  wood  or  stiff  wires  to  repre- 
sent lines  perpendicular  to  a  plane,  and  strings  to  represent  oblique  lines, 
any  figure  may  be  actually  made  with  a  comparatively  small  expenditure 
of  time  and  with  practically  no  expense.  For  reproductions  of  models 
actually  made  by  high  school  students,  see  group  on  p.  302  ;  also  §§  622, 
633,  678,  756,  762,  770,  797. 

606.  Assumption  20.     Revolution  postulate.      A  plane  may 
revolve  about  a  line  in  it  as  an  axis,  and  as  it  does  so  revolve,  it 
can  contain  any  particular  point  in  space  in  one  and  only  one 
position. 

607.  From  the  revolution  postulate  it  follows  that : 
Through  a  given  straight  line  any  number  of  planes  may  be 


For,  as  plane  MN  revolves  about  AB  as  an  axis  (§  606)  it 
may  occupy  an 
unlimited  num- 
ber of  positions 
each  of  which 
will  represent  a 
different  plane 
through  AB. 

608.  Def .     A  plane  is  said  to  be  determined  by  given  condi- 
tions if  that  pfane  and  no  other  plane  fulfills  those  conditions. 

609.  From  §§  607  and  608  it  is  seen  that: 
A  straight  line  does  not  determine  a  plane. 


Ex.  1153.  How  many  planes  may  be  passed  through  any  two  points 
in  space  ?  why  ? 

Ex.  1154.  At  a  point  P  in  a  given  straight  line  AB  in  space,  con- 
struct a  line  perpendicular  to  A  B.  How  many  such  lines  can  be  drawn  ? 


BOOK  VI 


301 


/    LINES   AND   PLANES 
**  PROPOSITION  I.     THEOREM 

610.  A  plane  is  determined  by  a  straight  line  and  a 
point  not  in  the  line. 

S 


A — - 


M 


R 


N 


^-B 


Given   line  AB  and  P,  a  point  not  in  AB. 
To  prove   that  AB  and  P  determine  a  plane. 


ARGUMENT 

1.  Through  AB  pass  any  plane,  as  MN. 

2.  Kevolve  plane  MN  about  AB  as  an  axis 

until  it   contains  point  P.     Call  the 
plane  in  this  position  RS. 

3.  Then   plane  RS  contains   line  AB   and 


REASONS 


1.  §  607. 

2.  §  606. 


3.    Arg.  2. 


point  P. 

4.  Furthermore,  in  no  other  position  can     4.    §  606. 

plane  MN,  in   its  rotation  about  AB, 
contain  point  P. 

5.  .•.  RS  is  the  only  plane  that  can  contain     5.    Arg.  4. 

AB  and  P. 

6.  .'.  AB  and  P  determine  a  plane.       Q.E.D.     6.    §  608. 

611.  Cor.  1. 1*  A  plane  is  determined  by  three  points 
in  the  same  straight  line. 

HINT.     Let  A,  B,  and  C  be  the  three  given  points.     Join  A  and  B  by  a 
straight  line,  and  apply  §  610. 

612.  Cor.  II.     A  plane  is  determined  by  two  intersec- 
ting straight  lines. 

1  '  *£  oo  oW 


not 


2/" 

.  ^ 

302  SOLID   GEOMETRY 

(  f 

613.  Cor.  III.  A  plane  is  determined  by  two  parallel 
straight  lines.  

Ex.  1155.  Given  line  AB  in  space,  and  P  a  point  not  in  AB.  Con- 
struct, through  P,  a  line  perpendicular  to  AB. 

Ex.  1156.  Hold  two  pencils  so  that  a  plane  can  be  passed  through 
them.  In  how  many  ways  can  this  be  done,  assuming  that  the  pencils  are 
lines?  why? 

Ex.  1157.  Can  two  pencils  be  held  so  that  no  plane  can  be  passed 
through  them  ?  If  so,  how  ? 

Ex.  1158.  In  measuring  wheat  with  a  half  bushel  measure,  the  meas- 
ure is  first  heaped,  then  a  straightedge  is  drawn  across  the  top.  Why  is 
the  measure  then  even  full  ? 

Ex.  1159.  Why  is  a  surveyor's  transit  or  a  photographer's  camera 
always  supported  on  three  legs  rather  than  on  two  or  four  ? 

Ex.  1160.  How  many  planes  are  determined  by  four  straight  lines, 
no  three  of  which  lie  in  the  same  plane,  if  the  four  lines  intersect :  (1)  at 
a  common  point  ?  (2)  at  four  different  points  ? 


614.  Def.     The  intersection  of  two  surfaces  is  the  locus  of 
all  points  common  to  the  two  surfaces. 

615.  Assumption  21.   Postulate.     Two  planes  having  one  point 
in  common  also  have  another  point  in  common. 


Reproduced  from  Models  made  by  High  School  Students 


BOOK   VI 


PROPOSITION  It.     THEOREM 


303 


If  two  planes  intersect,  their  intersection  is  a 
straight  line. 

S 

N 


Given   intersecting  planes  UN  and  RS. 

To  prove   the  intersection  of  MN  and  RS  a  str.  line. 


ARGUMENT 

1.  Let  A  and  B  be  any  two  points  common 

to  the  two  planes  MN  and  RS. 

2.  Draw  str.  line  AB. 

3.  Since   both  A  and  B  lie  in  plane  MN, 

str.  line  AB  lies  in  plane  MN. 

4.  Likewise  str.  line  AB  lies  in  plane  RS. 

5.  Furthermore   no   point   outside  of  AB 

can  lie  in  both  planes. 

6.  .•.  AB  is  the  intersection  of  planes  MN 

and  RS. 

7.  But  AB  is  a  str.  line. 

8.  .-.  the  intersection  of  MN  and  RS  is  a 

str.  line.  Q.E.D. 


REASONS 

1.  §  615. 

2.  §  54,  15. 

3.  §  603,  -a. 

4.  §  603,  a. 

5.  §  6J.O. 

6.  §  614. 

7.  Ar'g.2. 

.8.  Args.  6  and  7. 


Ex.  1161.     Is  it  possible  for  more  than  two  planes  to  intersect  in  a 
straight  line  ?    Explain. 

Ex.  1162.     By  referring  to  §§  26  and  608,  give  the  meaning  of  the 
expression,  "Two  planes  determine  a  straight  line." 

Ex.  1163.     Is  the  statement  in  Ex.  1162  always  true  ?     Give  reasons 
for  your  answer. 


304 


SOLID   GEOMETRY 


PROPOSITION  III.     THEOREM 


617.  If  three  planes,  not  passing  through  the  same 
line,  intersect  each  other,  their  three  lines  of  intersection 
are  concurrent,  or  else  they  are  parallel,  each  to  each. 


M 


N 


M 


FIG.  1. 


FIG.  2. 


Given   planes   MQ,  PS,  and  EN  intersecting   each   other   in 
lines  MN,  PQ,  and  ES-,  also  : 

I.    Given   MN  and  ES  intersecting  at  0  (Fig.  1). 
To  prove   MN,  PQ,  and  ES  concurrent. 


ARGUMENT 

1.  v  0  is  in  line  MN,  it  lies  in  plane  MQ. 

2.  v  0  is  in  line  ES,  it  lies  in  plane  PS. 

3.  .:  O,  lying  in  planes  MQ  and  PS,  must 

lie  in  their  intersection,  PQ. 

4.  .-.  PQ   passes   through  0;   i.e.  MN,  PQ, 

and  ES  are  concurrent  in  0.        Q.E.D. 

II.    Given    MN  II  ES  (Fig.  2). 
To  prove    PQ  II  MN  and  ES. 

ARGUMENT 

1.  PQ  and  MN  are  either  II  or  not  II. 

2.  Suppose  that  PQ  intersects  MN ;   then 

MN  also  intersects  ES. 

3.  But  this  is  impossible,  for  MN  II  ES. 

4.  .-.  PQ  II  MN. 

5.  Likewise  PQ  II  ES.  Q.E.D. 


REASONS 

1.  §  603,  a. 

2.  §  603,  a. 

3.  §  614. 

4.  Arg.  3. 


REASONS 

1.  §  161,  a. 

2.  §  617,  I. 

3.  By  hyp. 

4.  §  161,  b. 

5.  By  steps  sim- 

ilar to  1-4. 


BOOK  VI 


305 


Cor.    If  two  straight  lines  cure  parallel  to  a  third 
straight  line,  they  are  parallel  to  each  oilier. 


Given   lines  AB  and  CD,  each  II  EF. 
To  prove    AB  II  CD. 

ARGUMENT 

1.  AB  and  CD  are  either  ||  or  not  II. 

2.  Through  AB  and  EF  pass  plane  AF,  and 

through  CD  and  #F  pass  plane  CF. 

3.  Pass  a  third   plane   through   AB    and 

point  (7,  as  plane  BC. 

4.  Suppose  that  ^45  is  not  II  CD  ;  then  plane 

BC  will  intersect  plane  CF  in  some 
line  other  than  (7Z>,  as  OT. 

5.  Then  CH  II  ^. 

6.  But  CD  II  J^F. 

7.  .*.  Off  and    <7D,   two   straight    lines 

plane  CF,  are  both  II  EF. 

8.  This  is  impossible. 

9.  .'.  AB   II    CD.  Q.E.D 


n 


REASONS 

1.  §  161,  a. 

2.  §  613. 

3.  §  610. 

4.  §  613. 


5.  §  617,11. 

6.  By  hyp. 

7.  Args.  5  and  6. 

8.  §  178. 

9.  §  161,  6. 


619.  Def.      A  straight  line  is  perpendicular  {o  a  plane  if  it  is 

perpendicular   to   every   straight    line    in    the    plane   passing 
through  the  point  of  intersection  of  the  given  line  and  plane. 

620.  Def.     A  plane  is  perpendicular  to  a  straight  line  if  the 

line  is  perpendicular  to  the  plane.  ^ 

621.  Def.    If  a  line  is  perpendicular  to  a  plane,  its  point  of 
intersection  with  the  plane  is  called  the  foot  of  the  perpendicular. 


306 


SOLID   GEOMETRY 


PROPOSITION  IV.     THEOREM 

622&1  If  a  straight  line  Is  perpendicular  to  each  of 
two  Intersecting  straight  lines  at  their  point  of  Inter- 
section, It  Is  perpendicular  to  the  plane  of  those  lines. 


Given  str.  line  FBj_AB  and  to  BC  at  B,  and  plane  MN  con- 
taining AB  and  BC. 

To  prove   FB  J_  plane  MN. 

OUTLINE  OF  PROOF 

1.  In  plane  MN  draw  AC,  through  B  draw  any  line,  as  BH, 
meeting  AC  at  H. 

2.  Prolong  FB  to  E  so  that  BE  =  FB-,  draw  AF,  HF,  CF,  AE, 
HE,  CE. 

3.  AB  and  BC  are  then  _L  bisectors  of  FE ;    i.e.  FA  —  AE, 
FC  =  CE. 

4.  Prove  A  AFC  =  A  EAC ;  then  Z  JL4.F  =  Z  .EL4#. 

5.  Prove  A  HA F  —  A  EAH ;  then  HF  =  HE. 

6.  .-.BH^FE;  i.e.  FB±.BH,  any  line  in  plane  MN  passing 
through  B. 

7.  .'.  FBA.MN. 

623.  Cor.  All  the  perpendiculars  that  can  be  drawn 
to  a  straight  line  at  a  given  point  In  the  line  lie  in  a 
plane  perpendicular  to  the  line  at  the  given  point. 


BOOK  VI  307 


PROPOSITION  V.     PROBLEM 

624.   Through  a  given  point  to  construct  a  plane  per- 
pendicular to  a  given  line. 


FIG.  1.  FIG.  2. 

Given   point  P  and  line  AB. 

To  construct,    through  P,  a  plane  _L  AB. 

I.    Construction 

1.  Through  line  AB  and  point  P  pass  a  plane,  as  APD  (in 
Fig.  1,  any  plane  through  AB}.     §§  607,  610. 

2.  In  plane  APD  construct  PD,  through  P,  _L  AB.    §§  148, 149. 

3.  Through  AB  pass  a  second  plane,  as  ABC.     §  607. 

4.  In  plane  ABC,  through  the  foot  of  PD,  construct  a  J_  to  AB 
(PC  in  Fig.  1,  DC  in  Fig.  2).     §  148. 

5.  Plane  MN,  determined  by  C,  D,  and  P,  is  the  plane  required. 

II.  The  proof  is  left  as  an  exercise  for  the  student. 
HINT.     Apply  §  623". 

III.  The  discussion  will  be  given  in  §  625. 


Ex.  1164.     Tell  how  to  test  whether  or  not  a  flag  pole  is  erect. 

Ex.  1165.  Lines  AB  and  CD  are  each  perpendicular  to  line  EF. 
Are  AB  and  CD  necessarily  parallel  ?  Explain.  Do  they  necessarily  lie 
in  the  same  plane  ?  why  or  why  not  ? 


308  SOLID  GEOMETRY 

PROPOSITION  VI.     THEOREM 

625.   Through  a  given  point  there  exists  only  one  plane 
perpendicular  to  a  given  line. 


/  ---•""  ~7"          /  "--        / 

/ **  y    /* -1-  -y 

jf/ /  ML 1  f 

FIG.  1.  FIG.  2. 

Given   plane  MN9  through  P,  *1~AB. 
To  prove   3/JV  the  only  plane  through  P  _L  AS. 

ARGUMENT  ONLY 

1.  Either  MN  is  the  only  plane  through  P  J_  AB  or  it  is  not. 

2.  In  MN  draw  a  line  through  P  intersecting  Jine  A#,  as  PR. 

3.  Let  plane  determined  by  AB  and  Ptf  be  denoted  by  APR. 

4.  Suppose  that  there  exists  another  plane  through  P  J 4#; 

let  this  second  plane  intersect  plane  APR  in  line  PS. 

5.  Then  AB  J_  P#  and  also  PS;   i.e.  PR  and  PS  are  _L  ^/?. 

6.  This  is  impossible. 

7.  .-.  MN  is  the  only  plane  through  P  J_  ^5.  Q.E.D. 


626.  Question.      In  Fig.  2,  explain  why  AB  J_  P#. 

627.  §§  624  and  625  may  be  combined  in  one  statement: 
Through  a  given  point  there  exists  one  and  only  one  plane  per- 
pendicular to  a  given  line. 

628.  Cor.  I.     The  locus  of  all  points  in  space  equidis- 
tant from  the  extremities  of  a  straight  line  segment  is 
the  plane  perpendicular  to  the  segment  at  its  mid-point. 

629.  Def.     A  straight  line  is  parallel  to  a  plane  if  the  straight 
line  and  the  plane  cannot  meet. 

630.  Def.     A  straight  line  is  oblique  to  a  plane  if  it  is  neither 
perpendicular  nor  parallel  to  the  plane. 

631.  Def.     Two  planes  are  parallel  if  they  cannot  meet. 


BOOK  VI 


309 


/  PROPOSITION  VII.     THEOREM 

V  632.   Two  planes  perpendicular  to  the  same  straight 
line  are  parallel.  A 


M' 


R 


Given   planes  MN  and  RS,  each  _L  line  AB. 

To  prove    MN  II  #S. 

HINT.     Use  indirect  proof.     Compare  with  §  187. 

PROPOSITION  VIII.     THEOREM 

633.   If  a  plane  intersects  two  parallel  planes,  the  lines 
of  intersection  are  parallel. 


Given    II  planes  MN  and  RS,  and  any  plane  PQ  intersecting 
MN  and  RS  in  AB  and  CD,  respectively. 
To  prove    AB  II  CD. 
HINT.     Show  that  AB  and  CD  cannot  meet. 

634.   Cor.  I.     Parallel    lines    intercepted    between    the 
same  parallel  planes  are  equal.    (HINT.    Compare  with  §  234.) 
Ex.  1166.     State  the  converse  of  Prop.  VIII.     Is  it  true  ? 


SOLID  GEOMETRY 

PROPOSITION  IX.     THEOREM 

635.  //  two  angles,  not  in  the  same  plane,  have  their 
sides  parallel  respectively,  and  lying  on  the  same  side  of 
the  line  joining  their  vertices,  they  are  equal-* 


R 

Given  Z  ABC  in  plane  MN  and  Z  DEF  in  plane  RS  with  BA 
and  BC  II  respectively  to  ED  and  EF,  and  lying  on  the  same  side 
of  line  BE. 

To  prove    Z  ABC=  Z  DEF. 


1. 
2. 
3. 
4. 
5. 
6. 
7: 
8. 

9. 

10. 
11. 
12. 
13. 


ARGUMENT 

Measure  off  BA  =  ED  and  BC  =  EF. 
Draw  AD,  CF,  AC,  and  DF.  - 
BA  II  ED  and  BC  II  ^. 
Then  ADEB  and  C^J^B  are  /17. 
/.  AD  =  BE  and  CF  =  BE. 
.'.AD=CF. 

Also  ^£  II  BE  and  OF  II  #E. 
.'.  AD  II  Cy^. 

/.  ACFD  is  a  O. 


But  BA  =  ED  and  £C7=  EF. 
.'.A  ABC  =  A  DEF. 
.'.  Z  ABC=  Z 


Q.E.D. 


REASONS 
§  122. 
§  54,  15. 

3.  By  hyp. 

4.  §  240. 
§  232. 

§  54,  1. 
§  220. 
§  618. 
§  240. 
§232. 
Arg.  1. 
§  116. 
§  110. 


1. 

2. 


5. 

6. 

7. 

8. 

9. 
10. 
11. 
12. 
13. 


Ex.  1167.     Prove  Prop.  IX  if  the  angles  lie  on  opposite  sides  of  l!h\ 

*  It  will  also  be  seen  (§  645)  that  the  planes  of  these  angles  are  parallel. 


BOOK  VI 
PROPOSITION  X.     THEOREM 


311 


J  636.   If  one  of  two  parallel  lines  is  perpendicular  to  a 
plane,  the  other  also  is  perpendicular  to  the  plane. 


M 


Given   AB  II  CD  and  AB  _L  plane  MN. 
To  prove    CD  J_  plane  MN. 

ARGUMENT 

1.  Through  D  draw  any  line  in  plane  MN, 

as  DF. 

2.  Through  B  draw  BE  in  plane  JOT  II  DF. 

3.  Then  Z  ABE  =  Z  CZXF. 

4.  But  Z  .45.E  is  a  rt.  Z. 

.'.  Z  CD^  is  a  rt.  Z;  ie.  CD  _L  DF,  any 

line  in  plane  MN  through  D. 
.'.  CD  _L  plane  JfJV. 


5. 


6. 


REASONS 

1.  §  54,  15. 

2.  §  179. 

3.  §  635. 

4.  §  619. 

5.  §  54,  1. 

6.  §  619. 


Ex.  1168.     In  the  accompanying  diagram  AS  and  CD  lie  in  the  same 
plane.     Angle    CBA  =  35°,    angle    BCD  =  35°, 
angle  ABE  =  90°,  BE  lying  in  plane  MN.     Is 
CD  necessarily  perpendicular  to  plane  MN? 
Prove  your  answer. 

Ex.  1169.     Can  a  line  be  perpendicular  to 
each  of  two  intersecting  planes  ?    Prove. 

Ex.  1170.  If  one  of  two  planes  is  per- 
pendicular to  a  given  line,  but  the  other 
is  not,  the  planes  are  not  parallel. 

Ex.  1171.     If  a  straight  line  and  a  plane  are  each  perpendicular  to  the 
same  straight  line,  they  are  parallel  to  each  other. 


M 


812 


SOLID   GEOMETRY 


PROPOSITION  XI.     PROBLEM 

637.   Through  a  given  point  to  construct  a  line  per- 
pendicular to  a  given  plane- 


FIG.  1. 


FIG.  2. 


Given   point  P  and  plane  1T#. 

To  construpt,   through  P,  a  line  _L  plane  JfJ\T. 

I.    Construction 

1.  In  plane  MN  draw  any  convenient  line,  as  AB. 

2.  Through  P  construct  plane  PQ  1_  AB.     §  624. 

3.  Let  plane  PQ  intersect  plane  MN  in  CD.     §  616. 

4.  In  plane   P  Q  construct   a  line  through  P  J_  CD,  as   PR. 
\  148,  149. 

5.  P#  is  the  perpendicular  required. 


II.    Proof 


ARGUMENT 


1.  Through  the  foot  of  PR  (P  in  Fig.  1, 

R   in   Fig.   2)    in    plane    MN,   draw 
EF  II  AB. 

2.  AB  _L  plane  PQ. 

3.  .-.  EF  ±  plane  PQ. 

4.  .-.  EF±PR-,  i.e.  PR  J_  tfJ1. 

5.  But  PR  J_  (7Z). 

6.  .-.  PR  JL  plane  MN.  Q.E.D. 

III.    The  discussion  will  be  given  in  §  639. 


REASONS 


1.    §  179. 


2.  By  cons. 

3.  §  636. 

4.  §  619. 
5..  By  cons. 
6.    §  622. 


BOOK   VI 


313 


PROPOSITION  XII.     THEOREM 

\J  638.   Through  a  given  point  there  exists  only  one  line 
perpendicular  to  a  given  plane. 


Given   point  P  and  line  PJ,  through  P,  _L  plane  MN. 
To  prove   PA  the  only  line  through  P  J_  MN. 


ARGUMENT 

1.  Either  PA   is   the   only   line    through 

P  _L  MN  or  it  is  not. 

2.  Suppose     there     exists    another     line 

through  P  _L  MN,  as  PB  •  then  PA  and 
PB  determine  a  plane. 

3.  Let  this   plane  intersect   plane  MN  in 

line  CD. 

4.  Then  PA  and  PB,  two  lines  through  P 

and    lying   in   the   same   plane,   are 
_L  CD. 

5.  This  is  impossible. 

6.  .-.  PA  is  the  only  line  through  P  _1_  MN. 

Q.E.D. 


REASONS 


1.  §  161,  a. 

2.  §  612. 


3.    §  616. 


4.   §  619. 


5.  §§  62,  154. 

6.  §  161,  b. 


639.  §§  637  and  638  may  be  combined  in  one  statement  as 
follows : 

Through  a  given  point  there  exists  one  and  only  one  line  per- 
pendicular to  a  given  plane. 


Ex.  1172.     Find  the  locus  of  all  points  in  a  plane  that  are  equidistant 
from  two  given  points  not  lying  in  the  plane. 


314 


SOLID   GEOMETRY 


PROPOSITION  XIII.     THEOREM 

\^p40.   Two  straight  lines   perpendicular  to  the  same 
plane  are  parallel.    * 


M 


7 


Given   str.  lines  AB  and  CD  J_  plane  MN. 
To  prove   AB  II  CD. 

The  proof  is  left  as  an  exercise  for  the  student. 
HINT.     Suppose  that  AB  is  not  ||  CZ>,  but  that  some  other  line  through 
B,  as  BE,  is  II  CD.     Use  §  638. 

PROPOSITION  XIV.     THEOREM 

641.   If  a  straight  line  is  parallel  to  a  plane,  the  in- 
tersection of  the  plane  with  any  plane  passing  through 
the  given  line  is  parallel  to  the  given  line. 
Ar \B 


Given  line  AB  II  plane  MN,  and  plane  AD,  through  AB,  inter- 
secting plane  MN  in  line  CD. 

To  prove    AB  II  CD. 

The  proof  is  left  as  an  exercise  for  the  student. 

HINT.  Suppose  that  AB  is  not  II  CD.  Show  that  AB  will  then  meet 
plane  MN. 


BOOK   VI 


315 


642.   Cor.  I.     If  a  plane  intersects  one  of  two  parallel 
lines,    it   must,   if   suffi- 
ciently   extended,    inter- 
sect the  other  also. 

HINT.  Pass  a  plane  through 
AB  and  CD  and  let  it  intersect 
plane  MN  in  EF.  Now  if  MN 
does  not  intersect  CD,  but  is  II  to 
it,  then  EF  II  CD,  §  641.  Apply 
§  178. 


D 


643.   Cor.  II.    If  two  in- 
tersecting lines  are  each 
parallel  to  a  given   plane,  the  plane  of  these  lines  is 
parallel  to  the  given 
plane. 

HINT.  If  plane  MN, 
determined  by  AB  and  CD, 
is  not  I). to  plane  RS,  it  will 
intersect  it  in  some  line,  as 
EF.  What  is  the  relation 
of  EF  to  AB  and  CD  ? 


644.  Cor.    III.      Prob- 
lem.    Through  a  given     

point  to  construct : 

(a)  A  line  parallel  to  a  given  plane. 

(V)  A  plane  parallel  to  a  given  plane. 

HINT,  (a)  Let  A  be  a  point  outside  of  plane  MN.  Through  A  con- 
struct any  plane  intersecting  plane  MN  in  line  CD.  Complete  the  con- 
struction. 

645.  Cor.  IV.     If  two  angles,  not  in  the  same  plane, 
have  their  sides  parallel  respectively,  their  planes  are 
parallel.  

Ex.  1173.  Hold  a  pointer  parallel  to  the  blackboard.  Is  its  shadow 
on  the  blackboard  parallel  to  the  pointer  ?  why  ?  \ 

Ex.  1174.  Find  the  locus  of  all  straight  lines  passing  through  a  given 
point  and  parallel  to  a  given  plane. 


316 


SOLID   GEOMETRY 


J 


PROPOSITION  XV.     THEOREM 


646.  If  two  straight  lines  are  parallel,  a  plane  con- 
taining one  of  the  lines,  and  only  one,  is  parallel  to 
the  other.  - » 


Given    II  lines  AB  and  CD,  and  plane  MN  containing  CD. 
To  prove    plane  MN  II  AB. 


ARGUMENT 

1.  Either  plane  MN  is  II  A  B  or  it  is  not. 

2.  Suppose  MN  is  not  II  AB ;  then  plane  MN 

will  intersect  AB. 

3.  Then  plane  JOT  must  also  intersect  CD. 

4.  This  is  impossible,  for  MN  contains  CD. 

5.  .-.  plane  MN  II  AB.  Q.E.D. 


REASONS 

1.  §  161,  a. 

2.  §  629. 

3.  §  642. 

4.  By  hyp. 

5.  §  161,  6. 


647.    Cor.  I.     Problem.    Through  a  given  line  to  construct 
a  plane  parallel  to  another  given  line- 


HINT.     Through  E,  any  point  in  <7Z>,  construct  a  line  HK II  AB. 

648.  Cor.  II.  Problem.  Through  a  given  point  to  con- 
struct a  plane  parallel  to  any  two  given  straight  lines  in 
space. 


BOOK  VI 


317 


PROPOSITION  XVI.     THEOREM 

649.  If  two  straight  lines  are  intersected  by  three  par- 
allel planes,  the  corresponding  segments  of  these  lines 
are  proportional- 


Given    II  planes    MN,  PQ,  and   RS  intersecting   line   AB  in 
A,  E,  B  and  line  CD  in  (7,  H,  D,  respectively. 

AE      CH 

To  prove      —  =  —  • 
EB      HD 

ARGUMENT 

1.  Draw  AD  intersecting  plane  PQ  in  F. 

2.  Let  the  plane  determined  by  AB  and  AD 

intersect  PQ  in  EF  and  RS  in  Bf). 

3.  Let  the   plane  determined  by  AD  and 

DC  intersect  PQ  in  FH  and  MN  in  AC. 

4.  .-.  EF  II  BD  and  FH  II  AC. 


EB 

AE 
EB 


FD 
CH 

HD' 


HD       FD 


Q.E.D. 


REASONS 
§  54,  15. 
§§  612,  616. 


3.  §§612,  616. 

4.  §  633. 

5.  §  410. 

6.  §54,1. 


650.  Cor.  If  two  straight  lines  are  intersected  by  three 
parallel  planes,  the  lines  are  divided  proportionally. 

Ex.  1175.  If  any  number  of  lines  passing  through  a  common  point 
are  cut  by  two  or  more  parallel  planes,  their  corresponding  segments  are 
proportional. 

Ex.  1176.  In  the  figure  for  Prop.  XVI^  AE  =  6,  EB  =  8,  AD  =  21, 
CD  =  28.  Find  AF  and  HD. 


318  SOLID   GEOMETRY 


J 


PROPOSITION  XVII.     THEOREM 

651.   A  straight  line  perpendicular  to  one  of  two  par 
allel  planes  is  perpendicular  to  the  other  also. 

A 


B 

Given   plane  MN  II  plane  RS  and  line  AB  _L  plane  RS. 
To  prove   line  AB  J_  plane  MN. 

ARGUMENT  ONLY 

1.  In  plane  MN,  through  (7,  draw  any  line  (7Z),  and  let  the 
plane  determined  by  AC  and  CD  intersect  plane  RS  in  EF. 

2.  Then  CD  II  EF. 

3.  But  AB  J_  EF. 

4.  .•.  AB  J_  CD,  any  line  in  plane  MN  passing  through  C. 

5.  /.  line  AB  _L  plane  MN.  Q.E.D. 

652.  Cor.  I.     Through  a  given  point  there  exists  only 
one  plane  parallel  to  a  given  plane.    (HINT.    Apply  §§  638, 
644&,  651,  625.) 

653.  §§  6446  and  652  may  be  combined  in  one  statement : 
Through  a  given  point  there  exists  one  and  only  one  plane 

parallel  to  a  given  plane. 

654.  Cor.  II.     If  two  planes  are  each  parallel  to  a  third 
plane,  they  are  parallel  to  each  other.    (HINT.    See  §  180.) 

655.  Def.     The  projection  of  a  point  upon  a  plane  is  the  foot 
of  the  perpendicular  from  the  point  to  the  plane. 

656.  Def.     The  projection  of  a  line  upon  a  plane  is  the  locus 
of  the  projections  of  all  points  of  the  line  upon  the  plane. 


BOOK  VI 


319 


PROPOSITION  XVIII.     THEOREM 

657.   The  projection  upon  a  plane  of  a  straight  line 
not  perpendicular  to  the  plane  is  a  straight  line. 


Given   str.  line  AB  not  _L  plane  MN. 

To  prove   the  projection  of  AB  upon  MN  a  str.  line. 


ARGUMENT 

1.  Through  C,  any  point  in  AB,  draw  CD  _L 

plane  MN. 

2.  Let  the  plane  determined  by  AB  and  CD 

intersect  plane  MN  in  the  str.  line  EF. 

3.  From  H,  any  point  in  AB,  draw  HK,  in 

plane  AF,  II  CD. 

4.  Then  HK  _L  plane  MN. 

5.  .-.  K  is  the  projection  of  H  upon  plane. 

6. 

7. 


MN. 


EF  is  the  projection  of  AB  upon  plane 


MN. 

the  projection  of 
is  a  str.  line. 


upon  plane  MN 
Q.E.D. 


REASONS 


1.    §  639. 


2.  §§  612,  616. 

3.  §  179. 

4.  §  636. 

5.  §  655. 

6.  §  656. 

7.  Args.  2  and  6. 


Ex.    1177.     Compare  the  length  of  the  projection  of  a  line  upon  a 
plane  with  the  length  of  the  line  itself  : 
(a)  If  the  line  is  parallel  to  the  plane. 

(6)  If  the  line  is  neither  parallel  nor  perpendicular  to  the  plane, 
(c)  If  the  line  is  perpendicular  to  the  plane. 


320  SOLID  GEOMETRY 

PROPOSITION  XIX.     THEOREM 

658.    Of  all  oblique  lines  drawn  from  a  point  to  a  plane : 

I.    Those  having  equal  projections  are  equal- 
II.    Those  having  unequal  projections  are  unequal,  and 
the  one  having  the  greater  projection  is  the  longer. 

P 


N 


Given   line  PO  J_  plane  MN  and  : 

I.  Oblique  lines  PA  and  PB  with  projection  OA  =  projec- 
tion OB. 

II.  Oblique  lines  PA  and  PC  with  projection  OC >  projec- 
tion OA. 

To  prove  :   I.    PB  =  PA-   II.    PC  >  PA. 

The  proof  is  left  as  an  exercise  for  the  student. 

659.  Cor.  I.     (Converse  of   Prop.  XIX).     Of  all  oblique 
lines  drawn  from  a  point  to  a  plane : 

I.   Equal  oblique  lines  have  equal  projections. 
II.    Unequal  oblique  lines  have  unequal  projections,  and 
the  longer  line  has  the  greater  projection. 

660.  Cor.  II.     The  locus  of  a  point  in  space  equidistant 
from  all  points  in  the  circumference  of  a  circle  is  a 
straight  line  perpendicular  to  the  plane  of  the   circle 
and  passing  through  its  center. 

661.  Cor.  III.     The  shortest  line  from  a  point  to  a  given 
plane  is  tli^e  perpendicular  from  that  point  to  the  plane. 

662.  Def.     The  distance  from  a  point  to  a  plane  is  the  length 
of  the  perpendicular  from  the  point  to  the  plane. 


BOOK  VI  321 

663.  Cor.  IV.     Two    parallel    planes    are    everywhere 
equally  distant.    (HINT.    See  §  634.) 

664.  Cor.  V.     If  a  line  is  parallel  to  a  plane,  all  points 
of  the  line  are  equally  distant  from  the  plane. 

Ex.  1178.     In  the  figure  of  §  658,  if  PO  -  12  inches,  PA  =  15  inches, 
and  PC  =  20  inches,  find  OA  and  CA'. 

Ex.  1179.     Find  the  locus  of  all  points  in  a  given  plane  which  are  at 
a  given  distance  from  a  point  outside  of  the  plane. 

Ex.  1180.     By  applying  §  660,  suggest  a  practical  method  of  con- 
structing a  line  perpendicular  to  a  plane  : 
(a)  Through  a  point  in  the  plane  ; 
(6)  Through  a  point  not  in  the  plane. 

Ex.  1181.     Find  a  point  in  a  plane  equidistant  from  all  points  in  the 
circumference  of  a  circle  not  lying  in  the  plane. 

Ex.  1182.     Find  the  locus  of  all  points  equidistant  from  two  parallel 
planes. 

Ex.  1183.     Find  the  locus  of  all  points  at  a  given  distance  d  from  a 
given  plane  MN. 

Ex.  1184.     Find  the  locus  of  all  points  in  space  equidistant  from  two 
parallel  planes  and  equidistant  from  two  fixed  points. 

Ex.  1185.     A  line  and  its  projection  upon  a  plane  always  lie  in  the 
same  plane. 

Ex.  1186.     (a)  The  acute  angle  that  a  straight  line  makes  with  its  own 
projection  upon  a  plane  is  the  least  angle  j 

that    it    makes    with    any    line    passing  K 

through  its  foot  in  the  plane.  |\\ 

(/>)   With  what  line  passing  through  /       \~\\B~~E7 

its  foot  and  lying  in  the  plane  does  it  /      C1    \    )      —/ 

make  the  greatest  angle  ?  /•_  *#        / 

HINT,     (a)    Measure    off    BD  =  BC.      *f 
Which  is  greater,  AD  or  AC  ?    By  means  of  §  173,  prove  Z  ABC  <  ZABD. 


665.  Def.  The  acute  angle  that  a  straight  line,  not  perpen- 
dicular to  a  given  plane,  makes  with  its  own  projection  upon 
the  plane,  is  called  the  inclination  of  the  line  to  the  plane. 


Ex.  1187.     Find  the  projection  of  a  line  12  inches  long  upon  a  plane, 
if  the  inclination  of  the  line  to  the  plane  is  30°  ;  45° ;  60°. 


322  SOLID   GEOMETRY 

DIHEDRAL   ANGLES 

666.  Defs.    A  dihedral   angle  is  the  figure   formed   by  two 
planes  that  diverge  from  a  line.     The   planes   forming  a  di- 
hedral angle  are  called  --its  faces,  and  the  intersection  of  these 
planes,  its  edge. 

667.  A  dihedral   angle  may  be  designated   by   reading   in 
order  the  two  planes  forming  the  angle;  thus,  an  angle  formed 
by  planes  AB  and  CD  is  angle  AB-CD, 

and  is  usually  written  angle  A-BC-D.  If 
there  is  no  other  dihedral  angle  having 
the  same  edge,  the  line  forming  the  edge 
is  a  sufficient  designation,  as  dihedral 
angle  BC. 

668.  Def.     Points,   lines,   or   planes 
lying  in  the  same  plane  are  said  to  be 
coplanar. 

669.  A  clear  notion  of  the  magnitude 

of  a  dihedral  angle  may  be  obtained  by  imagining  that  its  two 
faces,  considered  as  finite  portions  of  planes,  were  at  first 
coplanar  and  that  one  of  them  has  revolved  about  a  line  com- 
mon to  the  two.  Thus  in  the  figure  we  may  imagine  face  CD 
first  to  have  been  in  the  position  of  face  AB  and  then  to  have 
revolved  about  BC  as  an  axis  to  the  position  of  face  CD. 

670.  Def.     The  plane  angle  of  a  dihedral  angle  is  the  angle 
formed  by  two  straight  lines,  one  in  each  face  of  the  dihedral 
angle,  perpendicular  to  its  edge  at  the  same  point.     Thus  if 
EF,  in  face  AB,  is  _L  BC  at  F,  and  FH,  in  face  CD,  is  _L  BC  at  F, 
then  Z  EFHis  the  plane -Z  of  the  dihedral  Z  A-BC-D. 


Ex.  1188.     All  plane  angles  of  a  dihedral  angle  are  equal. 

Ex.  1189.  Is  the  plane  of  angle  EFH  (§  667)  perpendicular  to  the 
edge  BC?  Prove.  State  your  result  in  the  form  of  a  theorem. 

Ex.  1190.  Is  Ex.  309  true  if  the  quadrilateral  is  a  quadrilateral  in 
space,  i.e.  if  the  vertices  of  the  quadrilateral  are  not  all  in  the  same 
plane  ?  Prove. 


LOOK   VI 


323 


671.  Def.     Two  dihedral  angles  are  adjacent  if  they  have 
a   common   edge   and    a    common    face    which    lies   between 
them  ;  thus  Z  A-BC-D  and 

Z  D-CB-E   are    adj.    dihe- 
dral Zs. 

672.  Def.     If  one  plane 
meets    another    so    as    to 
make   two    adjacent   dihe- 
dral angles  equal,  each  of 
these  angles  is  a  right 
dihedral    angle,    and 

the  planes  are  said 
to  be  perpendicular 
to  each  other.  Thus 
if  plane  HP  meets 
plane  LM  so  that 
dihedral  A  H-KL-M 
and  M-LK-N  are 
equal,  each  Z  is  a 
rt.  dihedral  Z,  and  planes  HP  and  LM  are  _L  to  each  other. 


Ex.  1191.  By  comparison  with  the  definitions  of  the  corresponding 
terms  in  plane  geometry,  frame  exact  definitions  of  the  following  terms  : 
acute  dihedral  angle ;  obtuse  dihedral  angle ;  reflex  dihedral  angle ; 
oblique  dihedral  angle ;  vertical  dihedral  angles ;  complementary  di- 
hedral angles  ;  supplementary  dihedral  angles  ;  bisector  of  a  dihedral 
angle  ;  alternate  interior  dihedral  angles ;  corresponding  dihedral  angles. 
Illustrate  as  many  of  these  as  you  can  with  an  open  book. 

Ex.  1192.     If  one  plane  meets  another  plane,  the  sum  of  the  two 

adjacent  dihedral  angles  is  two  right  dihedral  angles. 
HINT.     See  proof  of  §  65. 

Ex.  1193.     If  the  sum  of  two  adjacent  dihedral  angles  is  equal  to  two 
right  dihedral  angles,  their  exterior  faces  are  coplanar. 
HINT.     See  proof  of  §  76. 

Ex.  1194.  If  two  planes  intersect,  the  vertical  dihedral  angles  are 
equal. 

HINT.     See  proof  of  §  77. 


324 


SOLID   GEOMETRY 


PROPOSITION  XX.     THEOREM 

673.  If  two  dihedral  angles  are  equal,   their  plane 
angles  are  equal. 


C' 


Given   two  equal  dihedral  A  BO  and  B'c1  whose  plane  A  are 
A  MNO  and  M'N'O',  respectively. 
To  prove    Z.  MNO  =  Z. 


REASONS 
1.    §  54,  14. 


2.  §  670.  ' 

3.  §  62. 

4.  §§  670,  62. 

5.  §  18. 


674.  Cor.  I.     T&e  plane  angle  of  a  right  dihedral  angle 
is  a  right  angle. 

675.  Cor.  n.     If  two  intersecting  planes  are  each  per- 
pendicular to  a  third  plane,  their  intersections  with,  ////? 
third  plane  intersect  each  other. 


ARGUMENT 

1.  Superpose  dihedral  Z  BC  upon  its  equal, 

dihedral  Z  B'c',  so  that  point  N  of 
edge  BC  shall  fall  upon  point  N1  of 
edge  B'C'. 

2.  Then  MN  and  M'N1,  two  lines  in  plane 

AB,  are  _L  BC  at  point  N. 

3.  .'.  JfArand  M'N'  are  collinear. 

4.  Likewise  NO  and  JV'O'  are  collinear. 

5.  .'.Z  MNO  =  Z  M'N'O'.  Q.E.D. 


BOOK  VI 


325 


Given  planes  AB 
and  CD  J_  plane  MN 
and  intersecting  each 
other  in  line  DB ;  also 
let  AE  and  FC  be  the 
intersections  of  planes 
AB  and  CD  with  plane 
MN. 

To  prove  that  AE 
and  FC  intersect  each 
other. 

ARGUMENT 

1.  Either  ^  II  TO  or  AE  and  TO  intersect 

each  other. 

2.  Suppose  ^  II  FC.     Then  through  77,  any 

point  in  DB,  pass  a  plane  HKL  _L  TO, 
intersecting  FC  in  #  and  -4#  in  £. 

3.  Then  plane  HKL  is  _L  ^4#  also. 

4.  /.  Z  .HJ5TL  is  the  plane  Z  of  dihedral  Z 

TO,  and  Z  KLH  is  the  plane  Z  of  di- 
hedral Z  .4#. 

But  dihedral  A  FC  and  AE  are  rt.  dihe- 
dral A. 

.' .  A  HKL  and  -SXZT  are  rt.  A. 

. ' .  A  FJPL  contains  two  rt.  A. 

8.  But  this  is  impossible. 

9.  .'.  AE  and  FC  intersect  each  other.  Q.E.D. 


REASONS 

1.  161,  a. 

2.  §  627. 


§636. 
§  670. 


5.  §  672. 

6.  §  674. 

7.  Arg.  6. 

8.  §  206. 

9.  §  161,  6. 


Ex.  1195.  Find  the  locus  of  all  points  equidistant  from  two  given 
points  in  space. 

Ex.  1196.  Find  the  locus  of  all  points  equidistant  from  three  given 
points  in  space. 

Ex.  1197.  Are  the  supplements  of  equal  dihedral  angles  equal  ? 
complements  ?  Prove  your  answer. 

Ex.  1198.  If  two  planes  are  each  perpendicular  to  a  third  plane, 
can  they  be  parallel  to  each  other?  Explain.  If  they  are  parallel  to 
each  other,  prove  their  intersections  with  the  third  plane  parallel. 


326 


SOLID   GEOMETRY 


PROPOSITION  XXI.     THEOREM 
(Converse  of  Prop.  XX) 

\j  676.   If  the  plane  angles  of  two  dihedral  angles  are 
equal,  the  dihedral  angles  are  equal. 


C" 


Given   two  dihedral  A  EG  and  BfCf  whose  plane  A  MNO  and 
M'N'O'  are  equal. 

To  prove   dihedral  Z  BC  =  dihedral  Z  B'c'. 


ARGUMENT 


1.   Place    dihedral   Z  BC    upon    dihedral 


'  so  that  plane  Z.MNO  shall  be 
superposed  upon  its  equal,  plane 
Z  M'N'O'. 

2.  Then  BC  and  B'C'  are  both  _L  Jf#  and 

NO  at  ^. 

3.  .-.  BC  and  £'CY'  are  both  J_  plane  MNO 

at  AT. 

4.  .*.  .6(7  and  -B'C'  are  collinear. 


5.  .-.  planes  AB  and  A'B',  determined   by 

MN  and  BC,  are  coplanar  ;  also  planes 
CD  and  C'D',  determined  by  BC  and 
NO,  are  coplanar. 

6.  .-.  dihedral  Z.  BC  =  dihedral  Z  £'<?'. 

Q.E.D. 


KEASONS 


1.   §  54,  14. 


2.   §  670. 


3.    §  622. 


4.  §  638. 

5.  §  612. 


6.    §  18. 


BOOK  VI  327 

677.  Cor.  If  the  plane  angle  of  a  dihedral  angle  is  a 
right  angle,  the  dihedral  angle  is  a  right  dihedral 
angle.  

Ex.  1199.  Prove  Ex.  1194  by  applying  §  676. 

Ex.  1200.  If  two  parallel  planes  are  cut  by  a  transversal  plane,  the 
alternate  interior  dihedral  angles  are  equal. 

HINT.  Let  Z  ABC  be  the  plane  Z  of 
dihedral  Z  V-WX-Y.  Let  the  plane 
determined  by  AB  and  BC  intersect  plane 
MN  in  CD.  Then  AB  and  CD  lie  in 
the  same  plane  and  are  II  (§  633).  Prove 
that  Z  DCB  is  the  plane  Z  of  dihedral 
Z  M-ZY-X. 

Ex.  1201.  State  the  converse  of  Ex. 
1200,  and  prove  it  by  the  indirect  method. 

Ex.  1202.  If  two  parallel  planes  are 
cut  by  a  transversal  plane,  the  correspond- 
ing dihedral  angles  are  equal.  (HINT.  See  proof  of  §  190.) 

Ex.  1203.  State  the  converse  of  Ex.  1202,  and  prove  it  by  the  indirect 
method. 

Ex.  1204.  If  two  parallel  planes  are  cut  by  a  transversal  plane,  the 
sum  of  the  two  interior  dihedral  angles  on  the  same  side  of  the  transversal 
plane  is  two  right  dihedral  angles.  (HINT.  See  proof  of  §  192.) 

Ex.  1205.  Two  dihedral  angles  whose  faces  are  parallel,  each  to  each, 
are  either  equal  or  supplementary  dihedral  angles.  (HINT.  See  proof  of 
§  198.) 

Ex.  1206.  A  dihedral  angle  has  the  same  numerical  measure  as  its 
plane  angle.  (HINT.  Proof  similar  to  that  of  §  358.) 

Ex.  1207.  Two  dihedral  angles  have  the  same  ratio  as  their  plane 
angles. 

Ex.  1208.  Find  a  point  in  a  plane  equidistant  from  three  given 
points  not  lying  in  the  plane. 

Ex.  1209.  If  a  straight  line  intersects  one  of  two  parallel  planes,  it 
must,  if  sufficiently  prolonged,  intersect  the  other  also.  (HINT.  Use  the 
indirect  method  and  apply  §§  663  and  664.) 

Ex.  1210.  If  a  plane  intersects  one  of  two  parallel  planes,  it  must,  if 
sufficiently  extended,  intersect  the  other  also.  (HINT.  Use  the  indirect 
method  and  apply  §  652.) 


328 


SOLID   GEOMETRY 


PROPOSITION  XXII.     THEOREM 

678.  If  a  straight  line  is  perpendicular  to  a  plane, 
every  plane  containing  this  line  is  perpendicular  to  the 
given  plane. 


*l 

N 

B 

n      / 

M 


Given   str.  line  AS  J_  plane  MN  and  plane  PQ  containing  line 
AB  and  intersecting  plane  MN  in  CD. 
To  prove   plane  PQ  _L  plane  MN. 


ARGUMENT 

1.  AB  _L  CD. 

2.  Through  B,  in  plane  MN,  draw  BE  J_  CD. 

3.  Then  Z  ABE  is  the  plane  Z  of  dihedral 

Z  Q-CD-M. 

4.  But  Z  ABE  is  a  rt.  Z. 

5.  .-.  dihedral  Z  Q-C7>-,¥is  a  rt.  dihedral 

Z,  and  plane  PQ_L  plane  MJV.      Q.E.D. 


REASONS 
1.    §619. 


§63. 
§670. 


4.  §619. 

5.  §677. 


Ex.  1211.  If  from  the  foot  of  a  per- 
pendicular to  a  plane  a  line  is  drawn  at 
right  angles  to  any  line  in  the  plane,  the 
line  drawn  from  the  point  of  intersec- 
tion so  formed  to  any  point  in  the  per- 
pendicular is  perpendicular  to  the  line  of 
the  plane. 

HINT.     Make  KE  =  EH.    Prove  AK  -  AH,  and  apply  §  142. 

Ex.  1212.     In  the  figure  of  Ex.  1211,  if  AB  =  20,  BE  =  4  VTl,  and 
EK=  10,  find  AK. 


BOOK  VI 


329 


PROPOSITION  XXIII.     THEOREM 

\v679.  //'  two  planes  are  perpendicular  to  each  other, 
any  line  in  one  of  them,  perpendicular  to  their  intersec- 
tion, is  perpendicular  to  the  other. 


«* 

N 

B 

n     / 

E; 


M 


Given   plane  PQ  J_  plane  MN,  CD  their  line  of  intersection, 
and  AB,  in  plane  PQ,  _L  CD. 
To  prove   AB  _L  plane  MN. 


ARGUMENT 

1.  Through.  JS,  in  plane  MN,  draw  BEJLCD. 

2.  Then  Z  ABE  is  the  plane  Z  of  the  rt. 

dihedral  Z  Q-CD-M. 

3.  .-.  Z  J£#  is  a  rt.  Z,  and  ^41?  _L  BE. 

4.  But  .45  _L  CD. 

5.  .-.  ^£  _L  plane  MN.  Q.E.D. 


REASONS 

1.  §63. 

2.  §670. 

3.  §674. 

4.  By  hyp. 

5.  §622. 


680.  Cor.  If  two  planes  are  perpendicular  to  each 
other,  a  line  perpendicular  to  one  of  them  at  any  point 
in  their  line  of  intersection,  lies  in  the  other. 

HINT.     Apply  the  indirect  method,  using  §§  679  and  638. 


Ex.  1213.  If  a  plane  is  perpendicular  to  the  edge  of  a  dihedral  angle, 
is  it  perpendicular  to  each  of  the  faces  of  the  dihedral  angle  ?  Prove  your 
answer. 

Ex.  1214.  The  plane  containing  a  straight  line  and  its  projection 
upon  a  plane  is  perpendicular  to  the  given  plane. 

Ex.  1215.  If  two  planes  are  perpendicular  to  each  other,  a  line  per- 
pendicular to  one  of  them  from  any  point  in  the  other  lies  in  the  other 
plane. 


330 


SOLID   GEOMETRY 


PROPOSITION  XXIV.     THEOREM 

681.  If  each  of  two  intersecting  planes  is  perpendicu- 
lar to  a  third  plane; 

I.   Their  line  of  intersection  intersects  the  third  plane. 

II.   Their  line  of  intersection  is  perpendicular  to  the 

third  plane. 


M 


Given   planes  PQ  and  RS  _L  plane  MN  and  intersecting  each 
other  in  line  AB. 

To  prove :       I.   That  AB  intersects  plane  MN. 
II.   AB  J_  plane  MN. 


I. 


ARGUMENT 


1.  Let  planes  PQ  and  RS  intersect  plane  MN  in 

lines  PD  and  RE,  respectively. 

2.  Then  PD  and  RE  intersect  in  a  point  as  C. 

3.  .•.  AB  passes   through   G\    i.e.  AB  intersects 

plane  MN.  Q.E.D. 


II. 


ARGUMENT 


1.  Either  AB  _L  plane  MN  or  it  is  not. 

2.  Suppose  AB  is  not  J_  plane  MN,  but  that  some 

other  line  through  <7,  the  point  common  to 
the  three  planes,  is  _L  plane  MN,  as  line  OF. 

3.  Then  CF  lies  in  plane  PQ,  also  in  plane  RS. 

4.  .-.  CF  is  the  intersection  of  planes  PQ  and  RS. 

5.  .-.  planes  PQ  and  RS  intersect  in  two  str.  lines, 

which  is  impossible. 
'fe.   .-.  AB  _L  plane  MN.  Q.E.D. 


REASONS 

1.  §616. 

2.  §675. 

3.  §  617,  I. 

REASONS 

1.  §161,  a. 

2.  §  639. 


3.  §680. 

4.  §  614. 

5.  §616. 

6.  §  161,  b. 


BOOK  VI 


331 


PROPOSITION  XXV.     PROBLEM 

682.  Through  any  straight  line,  not  perpendicular  to 
a  plane,  to  construct  a  plane  perpendicular  to  the  given 
plane. 


Given   line  AB  not  _L  plane  MN. 
To  construct,   through  AB,  a  plane  _L  plane  MN. 
The  construction,  proof,  and  discussion  are  left  as  an  exer- 
cise for  the  student. 

HINT.     Apply  §  678.    For  discussion,  see  §  683. 

683.  Cor.     Through  a  straight  line,  not  perpendicular 
to  a  plane,  there  exists  only  one  plane  perpendicular  to 
the  given  plane. 

HINT.     Suppose  there  should  exist  another  plane  through  AB  _L  plane 
MN.     What  would  you  know  about  AB  ? 

684.  §§682  and  683  may  be  combined  in  one  statement  as 
follows  : 

Through  a  straight  line,  not  perpendicular  to  a  plane,  there 
exists  one  and  only  one  plane  perpendicular  to  the  given  plane. 


Ex.  1216.  Apply  the  truth  of  Prop.  XXIV :  (a)  to  the  planes  that 
intersect  at  the  corner  of  a  room  ;  (6)  to  the  planes  formed  by  an  open 
book  placed  perpendicular  to  the  top  of  the  desk. 

Ex.  1217.  If  a  plane  is  perpendicular  to  each  of  two  intersecting 
planes,  it  is  perpendicular  to  their  intersection. 


332 


SOLID   GEOMETRY 


PROPOSITION  XXVI.     PROBLEM 

685.   To  construct  a  common  perpendicular  to  any  two 
'straight  lines  in  space. 

K  X 


Given   AB  and  CD,  any  two  str.  lines  in  space. 
To  construct   a  line  _L  both  to  AB  and  to  CD. 

I.    Construction 

1.  Through  CD  construct  plane  MN  II  AB.     §  647. 

2.  Through  AB  construct  plane  AF  J_  plane  MN  intersecting 
MN  in  EF,  and  CD  in  H.     §  682. 

3.  Through  H  construct  HK,  in  plane  AF,  _L  EF.     §  148. 

4.  HK  is  _L  to  both  AB  and  C7)  and  is  the  line  required. 


II.    Proof 


ARGUMENT 

1.  ^.S  II  .plane  MN. 

2.  .-.  tfF  II  AS. 

3.  But  #Jf  _L  EF. 

4.  .-.  /r.g'J.  AB. 

5.  Also  JMT  ±  plane  MN. 

6.  .-.  HK  ±_  CD. 

7.  .:  HK  _L  to  both  J£  and 


Q.E.D. 


REASONS 

1.  By  cons. 

2.  §  641. 

3.  By  cons. 

4.  §  193. 

5.  §  679. 

6.  §  619. 

7.  Args.  4  and  6. 


III.   The  discussion  will  be  given  in  §  686. 

686.  Cor.  Between  two  straight  lines  in  space  (not 
in  the  same  plane}  there  exists  only  one  common  per- 
pendicular. 


BOOK  VI  333 

HINT.  Suppose  XF,  in  figure  of  §  685,  a  second  _L  to  AB  and  CD. 
Through  T  draw  ZW  \\  AB.  What  is  the  relation  of  XT  to  AB?  to 
ZW?  to  CD?  to  plane  MN?  Through  X  draw  XE  _L  EF.  What  is 
the  relation  of  XB  to  plane  MN  ?  Complete  the  proof. 

687.  §§  685  and  686  may  be  combined  in  one  statement  as 
follows : 

Between  two  straight  lines  in  space  (not  in  the,  same  plane) 
there  exists  one  and  only  one  common  perpendicular. 


Ex.  1218.  A  room  is  20  feet  long,  15  feet  wide,  and  10  feet  high. 
Find  the  length  of  the  shortest  line  that  can  be  drawn  on  floor  and  walls 
from  a  lower  corner  to  the  diagonally  opposite  corner.  Find"  the  length  of 
the  line  that  extends  diagonally  across  the  floor,  then  along  the  intersec- 
tion of  two  walls  to  the  ceiling. 

Ex.  1219.  If  two  equal  lines  are  drawn  from  a  given  point  to  a  given 
plane,  the  inclinations  of  these  lines  to  the  given  plane  are  equal.  If  two 
unequal  lines  are  thus  drawn,  which  has  the  greater  inclination  ?  Prove. 

Ex.  1220.  The  two  planes  determined  by  two  parallel  lines  and  a 
point  not  in  their  plane,  intersect  in  a  line  which  is  parallel  to  each  of  the 
given  parallels. 

Ex.  1221.  If  two  lines  are  parallel,  their  projections  on  a  plane  are 
either  the  same  line,  or  parallel  lines. 

Ex.  1222.  If  each  of  three  planes  is  perpendicular  to  the  other  two  : 
(a)  the  intersection  of  any  two  of  the  planes  is  perpendicular  to  the  third 
plane  ;  (6)  each  of  the  three  lines  of  intersection  is  perpendicular  to  the 
other  two.  Find  an  illustration  of  this  exercise  in  the  classroom. 

Ex.  1223.  If  two  planes  are  parallel,  no  line  in  the  one  can  meet 
any  line  in  the  other. 

Ex.  1224.  Find  all  points  equidistant  from  two  parallel  planes  and 
equidistant  from  three  points  :  (a)  if  the  points  lie  in  neither  plane  ; 
(&)  if  the  points  lie. in  one  of  the  planes. 

Ex.  1225.  Find  all  points  equidistant  from  two  given  points,  equi- 
distant from  two  parallel  planes,  and  at  a  given  distance  d  from  a  third 
plane. 

Ex.  1226.  If  each  of  two  intersecting  planes  is  parallel  to  a  given 
line,  the  intersection  of  the  planes  is  parallel  to  the  line. 

Ex.  1227.  Construct,  through  a  point  in  space,  a  straight  line  that 
shall  be  parallel  to  two  intersecting  planes. 


334 


SOLID   GEOMETRY 


PROPOSITION  XXVII.     THEOREM 

688.  Every  point  in  the  plane  that  bisects  a  dihedral 
angle  is  equidistant  from  the  faces  of  the  angle. 


M 


Given  plane  BE  bisecting  the  dihedral  Z  formed  by  planes 
AG  and  CD ;  also  PH  and  PK  Js  from  P,  any  point  in  plane  BE, 
to  faces  AC  and  CD,  respectively. 

To  prove    PH=  PK. 


ARGUMENT 

1.  Through  PH  and  PK  pass  plane  MN 

intersecting  plane  AC  in  CH,  plane 
CD  in  CK,  plane  BE  in  PC,  and  edge 
BC  in  C. 

2.  Then  plane  MN  J_  planes  AC  and  (77); 

t.e.  planes  AC  and  CZ)  are  _L  plane  JOT. 

3.  /.   .BC  _L  plane  MN. 

4.  .-.  BC±  CH,  CP,  and  C#. 

5.  .-.  APCH  and  J5TCP  are  the  plane  A  of 

the  dihedral  A  E-BC-A  and  D-CB-E. 

6.  But    dihedral    Z  E-BC-A  =  dihedral 

Z  D-CB-E. 
7. 
8. 
9. 
10.   .\  PH  =  PIT.  Q.E.D. 


Also  PC  =  PC. 

.'.  rt.  A  PC/f=  rt.  A  J5TCP. 

.\  PH  =  PIT. 


REASONS 
1.    §  612,  616. 


2.  §  678. 

3.  §  681,  II. 

4.  §  619. 

5.  §  670. 

6.  By  hyp. 

7.  §  673. 

8.  By  iden. 

9.  §  209. 
10.  §  110. 


BOOK  VI  335 

689.  Cor.  I.    Every  point   equidistant  from  the   two 
faces  of  a  dihedral  angle  lies  in  the  plane  bisecting  the 
angle. 

690.  Cor.  n.     The  plane  bisecting  a  dihedral  angle  is 
the  locus  of  all  points  in  space  equidistant  from  the. 
faces  of  the  angle. 

691.  Cor.   m.     Problem.     To  construct  the  bisector  of  a 
given  dihedral  angle. 


Ex.  1228.  Prove  that  a  dihedral  angle  can  be  bisected  by  only  one 
plane. 

HINT.     See  proof  of  §  599. 

Ex.  1229.  Find  the  locus  of  all  points  equidistant  from  two  inter- 
secting planes.  Of  how  many  planes  does  this  locus  consist  ? 

Ex.  1230.  Find  the  locus  of  all  points  in  space  equidistant  from  two 
intersecting  lines.  Of  how  many  planes  does  this  locus  consist  ? 

Ex.  1231.  Find  the  locus  of  all  points  in  space  equidistant  from  two 
parallel  lines. 

Ex.  1232.  Find  the  locus  of  all  points  in  space  equidistant  from  two 
intersecting  planes  and  equidistant  from  all  points  in  the  circumference 
of  a  circle. 

Ex.  1233.  Find  the  locus  of  all  points  in  space  equidistant  from  two 
intersecting  planes  and  equidistant  from  two  fixed  points. 

Ex.  1234.  Find  the  locus  of  all  points  in  space  equidistant  from  two 
intersecting  planes,  equidistant  from  two  parallel  planes,  and  equidistant 
from  two  fixed  points. 

Ex.  1235.  If  from  any  point  within  a  dihedral  angle  lines  are  drawn 
perpendicular  to  the  faces  of  the  angle,  the  angle  formed  by  the  perpen- 
diculars is  supplementary  to  the  plane  angle  of  the  dihedral  angle. 

Ex.  1236.  Given  two  points,  P  and  Q,  one  in  each  of  two  intersecting 
planes,  M  and  N.  Find  a  point  JTin  the  intersection  of  planes  Mand  N 
such  that  PX+XQ  is  a  minimum. 

Ex.  1237.  Given  two  points,  P  and  Q,  on  one  side  of  a  given  plane 
MN.  Find  a  point  X  in  plane  MN  such  that  PX  -f  XQ  shall  be  a 
minimum. 

HINT.     See  Ex.  176. 


336  SOLID   GEOMETRY 

POLYHEDRAL    ANGLES 

692.  Def.  A  polyhedral  angle  is  the  figure  generated  by  a 
moving  straight  line  segment  that  continually  intersects  the 
boundary  of  a  fixed  polygon  and  one  extremity  of  which  is  a 
fixed  point  not  in  the  plane  of  the  given  polygon.  A  poly- 
hedral angle  is  sometimes  called  a  solid  angle. 

? 

693-  Defs.  The  moving  line  is  called  the 
generatrix,  as  V*A  ;  the  fixed  polygon  is  calLed 
the  directrix,  as  polygon  ABODE',  the  fixed 
point  is  called  the  vertex  of  the  polyhedral- 
angle,  as  F. 

694.  Defs.    The  generatrix  in  any  position 
is  an  element  of  the  polyhedral  angle ;  the 
elements  through  the  vertices  of  the  poly- 
gon .  are    the   edges,   as    VA,    VB,  etc. ;    the 
portions  of  the   planes  determined   by  the 

edges  of  the  polyhedral  angle,  and  limited  by  them  are  the 
faces,  as  AVB,  BVC,  etc.;  the  angles  formed  by  the  edges  are 
the  face  angles,  as  A  AVB,  BVC,  etc.;  the  dihedral  angles 
formed  by  the  faces  are  called  the  dihedral  angles  of  the  poly- 
hedral angle,  as  dihedral  A  VA,  VB,  etc. 

695.  Def.     The  face  angles  and  the  dihedral  angles  taken 
together  are  sometimes  called  the  parts  of  a  polyhedral  angle. 

696.  A  polyhedral  angle  may  be  designated  by  a  letter  at 
the  vertex  and  one  on  each  edge,  as  V-ABCDE.     If  there  is  no 
other  polyhedral -angle  having  the  same  vertex,  the  letter  at 
the  vertex  is  a  sufficient  designation,  as  V. 

697.  Def.     A  convex  polyhedral  angle  is  a  polyhedral  angle 
whose  directrix  is  a  convex  polygon,  i.e.  a  polygon  no  side  of 
which,  if  prolonged,  will  enter  the  polygon;  as  V-ABCDE.     In 
this  text  only  convex  polyhedral  angles  will  be  considered. 

698.  Defs.      A  trihedral  angle  is  a  polyhedral  angle  whose 
directrix  is  a  triangle  (£ri-gon) ;  a  tetrahedral  angle,  a  polyhe- 
dral angle  whose  directrix  is  a  quadrilateral  (tetra-gou)  ;  etc. 


BOOK   VI  337 

699.  Defs.     A  trihedral  angle  is  called  a  rectangular,  birec- 
tangular, or  trirectangular  trihedral  angle  according  as  it  contains 
one,  two,  or  three  right  dihedral  angles. 

700.  Def.    An  isosceles  trihedral  angle  is  a  trihedral  angle 
having  two  face  angles  equal. 


Ex.  1238.  By  holding  an  open  book  perpendicular  to  the  desk,  illus- 
trate birectangular  and  trirectangular  trihedral  angles.  By  placing  one 
face  of  the  open  book  on  top  of  the  desk  and  the  other  face  along  the  side 
of  the  desk  against  the  edge,  illustrate  a  rectangular  trihedral  angle. 

Ex.  1239.  Is  every  birectangular  trihedral  angle  isosceles  ?  Is  every 
isosceles  trihedral  angle  birectangular  ? 


701.  From  the  general  definition  of  equal  geometric  figures 
(§  18)  it  follows  that: 

Two  polyhedral  angles  are  equal  if  they  can  be  made  to  coincide. 

PROPOSITION  XXVIII.     THEOREM 

702.  Two  trihedral  angles  are  equal  : 

I.  If  a  face  angle  and  the  two  adjacent  dihedral  an- 
gles of  one  are  equal  respectively  to  a  face  angle  and 
tlie  two  adjacent  dihedral  angles  of  tfo  other; 

II.  If   two   face    angles   and   the  included  dihedral 
angle  of  one  are  equal  respectively  to  two  face  angles 
and  the  included  dihedral  angle  of  the  other: 
provided  the  equal  parts  are  arranged  in  the  same  order. 

The  proofs  are  left  as  exercises  for  the  student. 

703.  Questions.      Compare    care- 
fully  the  wording  of  I  above  and  the 
accompanying  figures  with  the  wording 
and  figures  of  §  105.     What  in  I  takes 
the  place  of  A  in  §  105?  side?  adj.  A? 
What,  in  the  accompanying  figure,  cor- 
responds to  A   ABC  in  the  proof  of 
§105?    A    DEF?    AC?    DF?    point 

A?  point  C?    If  these  and  similar  changes  are  made  in  the  proof  of 
§  105,  will  it  serve" as  a  proof  of  I  above  ?    Compare  II  above  with  §  107. 


338 


SOLID   GEOMETRY 


PROPOSITION  XXIX.     THEOREM 

704.  Two  trihedral  angles  are  equal  if  the  three  face 
angles  of  one  are  equal  respectively  to  the  three  face 
angles  of  the  other,  and  the  equal  parts  are  arranged  in 
the  same  order. 


Given  trihedral    A  V-ABC  and  v'-A'B'c',  Z  A  VB  =  Z  A' 
Z  BVC  =  Z  B'V'C',  Z  CVA  =  Z  c'v'A',  and  the  equal  face  angles 
arranged  in  the  same  order. 

To  prove   trihedral  Z  V-ABC=  trihedral  Z  v'-A'&C*. 

OUTLINE  OF  PROOF 

1.  Since,  by  hyp.,  any  two  face  A  of   V-ABC,  as  A  AVB 
and  BVC,  are  equal,  respectively,  to  the  two  corresponding  face 
A  of  v'-A'B'C1,  it  remains  only  to  prove  the  included  dihe- 
dral A  VB  and  V'B'  equal.     §  702,  II.     (See  also  §  705.) 

2.  Let  face  A  AVB  and  BVC  be  oblique  A]  then  from  any 
point  E  in  VB,  draw  ED  and  EF,  in  planes  AVB  and  BVC,  respec- 
tively, and  _L  VB. 

3.  Since  A  AVB  and  BVC  are  oblique  A,  ED  and  EF  will 
meet  VA  and  VC  in  D  and  F,  respectively.     Draw  FD. 

4.  Similarly,  lay  off  V'E'  =  VE  and  draw  A  D'E'F'. 

5.  Prove  rt.  &DVE=rt.  AD1  V'E';  thenVD=v'D',ED  =  E'D'. 
6;   Prove  rt.  A  EVF=  rt.  A  E'V'F';  then  VF=  V'F',  EF=E'F'. 

7.  Prove  A  FVD  =  A  F'V'D1 ;  then  FD  =  F'D'. 

8.  .  • .  A  DEF  =  A  D'E'F' ;  then  Z  ZXE^  =  Z 


BOOK  VI  339 

9.   But  A  DEF  and  D'E'F'  are  the  plane  A  of  dihedral  A  VB 
and  V'B',  respectively. 

10.  .-.  dihedral  Z  VB  =  dihedral  Z  V'B1. 

11.  .-.  trihedral  Z  v -AB C  =  trihedral  Z  V'-A'B'C'.         Q.E.D. 

705.  Note.     If  all  the  face  A  are  rt.  A,  show  that  all  the  dihedral  A 
are  rt.  dihedral  A  and  hence  that  all  are  equal.     If  two  face  A  of  a  trihe- 
dral Z  are  rt.  A,  show  that  the  third  face  Z  is  the  plane  Z  of  the  included 
dihedral  Z,  and  hence  that  two  homologous  dihedral  A,  as  VB  and  V'B1, 
are  equal.     It  remains  to  prove  that  Prop.  XXIX  is  true  if  only  one  face  Z 
of  the  first  trihedral  Z  and  its  homologous  face  Z  of  the  other  are  rt.  A, 
or  if  all  face  A  are  oblique. 

706.  Questions.     State  the  proposition  in  Bk.  I  that  corresponds  to 
§  704.     What  was  the  main  step  in  the  proof  of  that  proposition?    Did 
that  correspond  to  proving  dihedral  Z  VB  of  §  704  =  dihedral  Z  V'B'  ? 

707.  Def.     Two    polyhedral    angles   are   said    to   be   sym- 
metrical if  their  corresponding  parts  are  equal  but  arranged  in 
reverse  order. 

By  making  symmetrical  polyhedral  angles  and  comparing 
them,  the  student  can  easily  satisfy  himself  that  in  general 
they  cannot  be  made  to  coincide. 

708.  Def.     Two  polyhedral  angles  are  said  to  be  vertical 
if  the  edges  of  each  are  the  prolongations  of  the  edges  of  the 
other. 

It  will  be  seen  that  two  vertical,  like  two  symmetrical,  poly- 
hedral angles  have  their  corresponding  parts  equal  but 
arranged  in  reverse  order. 


Two  Equal  Polyhedral  Two  Vertical  Poly-  Two  Symmetrical  Poly- 

Angles  hedral  Angles  hedral  Angles 


340  SOLID   GEOMETRY 

PROPOSITION    XXX.     THEOREM 

709.   Two  trihedral  angles  are  symmetrical: 

I.  If  a  face  angle  and  the  two  adjacent  dihedral  an- 
gles of  one  are  equal  respectively  to  a  face  angle  and 
the  two  adjacent  dihedral  angles  of  the  other; 

II.  If  two  face   angles  and    the    included    dihedral 
angle  of  one  are  equal  respectively  to  two  face  angles 
and  the  included  dihedral  angle  of  the  other; 

III.  If  the  three  face  angles  of  one  are  equal  respec- 
tively to  the  three  face  angles  of  the  other  : 

provided  the  equal  parts  are  arranged  in  reverse  order- 


\ 
\ 


The  proofs  are  left  as  exercises  for  the  student. 

HINT.  Let  V  and  V  be  the  two  trihedral  A  with  parts  equal  but 
arranged  in  reverse  order.  Construct  trihedral  Z  V"  symmetrical  to  F. 
Then  what  will  be  the  relation  of  V"  to  V  ?  of  V  to  F? 


Ex.  1240.  Can  two  polyhedral  angles  be  symmetrical  and  equal  ? 
vertical  and  equal  ?  symmetrical  and  vertical  ?  If  two  polyhedral  angles 
are  vertical,  are  they  necessarily  symmetrical  ?  if  symmetrical,  are  they 
necessarily  vertical  ? 

Ex.  1241.  Are  two  trirectangular  trihedral  angles  necessarily  equal? 
Are  two  birectangular  trihedral  angles  equal  ?  Prove  your  answers. 

Ex.  1242.  If  two  trihedral  angles  have  three  face  angles  of  one  equal 
respectively  to  three  face  angles  of  the  other,  the  dihedral  angles  of  the 
first  are  equal  respectively  to  the  dihedral  angles  of  the  second. 


BOOK  VI  841 


PROPOSITION  XXXI.     THEOREM 

710.   The  sum  of  any  two  face  angles  of  a  trihedral 
angle  is  greater  than  the  third  face  angle- 


V 


c 

Given  trihedral  Z  V-ABC  in  which  the  greatest  face 
Z  is  A  vs. 

To  prove    Z  BVC  +  Z  CVA  >  Z  ^IFB. 
OUTLINE  OF  PROOF 

1.  In    face   AVB    draw    FZ)   making  Z  D  Ffi  =  Z  £  F(7,  and 
through  D  draw  any  line  intersecting  VA  in  #  and  VB  in  .F. 

2.  On  VC  lay  off  F£  =  VD  and  draw'^G  and  GE. 

3.  Prove  A  ^F£  =  A  DVF-,  then  ^  =  FD. 

4.  But  FG  +  GE  >  FD  +  DE ;   .'.  GE  >  £#. 

5.  In  A  GVE  and  ^FD,  pdft^£*«?Z  >  Z  J?FD. 

6.  But  Z  ^F£  =  ^LDVF. 

1.    .'.  Z  ^F£  +  Z  GVE  >Z.EVD  +  Z.D%F; 

i.e.  /.  BVC  +  Z.  CVA  >  Z  ^FB.  Q.E.D. 

711.  Question.  State  the  theorem  in  Bk.  I  that^orrespouds  to 
Prop.  XXXI.  Can  that  theorem  be  proved  by  a  method  similar  to  the 
one  used  here  ?  If  so,  give  the  proof. 


Ex.  1243.  If,  in  trihedral  angle  V-ABC,  angle  SVC  =  60°,  and  angle 
CVA  =  80°,  make  a  statement  as  to  the  number  of  degrees  in  angle  AVB. 

Ex.  1244.  Any  face  angle  of  a  trihedral  angle  is  greater  than  the 
difference  of  the  other  two. 


342  SOLID   GEOMETRY 


PROPOSITION  XXXII.     THEOREM 

712.   The  sum  of  all  the  face  angles  of  any  convex  poly- 
hedral angle  is  less  than  four  right  angles- 

V 


Given   polyhedral  Z  V  with  n  faces. 

To  prove   the  sum  of  the  face  A  at  V  Z  4  rt.  A. 

HINT.  Let  a  plane  intersect  the  edges  of  the  polyhedral  Z.  in  A,  B,  (T, 
etc.  From  O,  any  point  in  polygon  ABC  .  .  .,  draw  OA,  OB,  OC,  etc. 
How  many  A  have  their  vertices  at  V?  at  O  ?  What  is  the  sum  of 
all  the  A  of  all  the  A  with  vertices  at  V  ?  at  0  ?  Which  is  the  greater, 
Z  ABV  +  Z  VBG  or  Z  ABO  +  Z  OBC  ?  Then  which  is  the  greater,  the 
sum  of  the  base  A  of  A  with  vertices  at  T7,  or  the  sum  of  the  base  A  of  A 
with  vertices  at  O  ?  Then  which  is  greater,  the  sum  of  the  face  A  about 
T,  or  the  sum  of  the  A  about  0  ? 

713.  Question.  Is  there  a  proposition  in  plane  geometry  corre- 
sponding to  Prop.  XXXII  ?  If  so,  state  it.  If  not,  state  the  one  that 
most  nearly  corresponds  to  it. 


Ex.  1245.  Can  a  polyhedral  angle  have  for  its  faces  three  equi- 
lateral triangles  ?  four  ?  five  ?  six  ? 

Ex.  1246.  Can  a  polyhedral  angle  have  for  its  faces  three  squares  ? 
four  ? 

Ex.  1247.  Can  a  polyhedral  angle  have  for  its  faces  three  regular 
pentagons  ?  four  ? 

Ex.  1248.  Show  that  the  greatest  number  of  polyhedral  angles  that 
can  possibly  be  formed  with  regular  polygons  as  faces  is  five. 

Ex.  1249.  Can  a  trihedral  angle  have  for  its  faces  a  regular  decagon 
and  two  equilateral  triangles?  a  regular  decagon,  an  equilateral  tri- 
angle, and  a  square  ?  two  regular  octagons  and  a  square  ? 


BOOK   VII 

POLYHEDRONS 

714.  Def.     A  surface  is  said  to  be  closed  if  it  separates  a 
finite  portion  of  space  from  the  remaining  space. 

715.  Def.    A  solid  closed  figure  is  a  figure  in  space  composed  of 
a  closed  surface  and  the  finite  portion  of  space  bounded  by  it. 

716.  Def.     A   polyhedron   is   a   solid   closed   figure   whose 
bounding  surface  is  composed  of  planes  only. 

717.  Defs.    The  intersections  of  the 
bounding  planes  are  called  the  edges ; 
the   intersections   of  the   edges,   the 
vertices;    and    the    portions    of    the 
bounding  planes  bounded  by  the  edges, 
the  faces,  of  the  polyhedron. 

718.  Def.     A  diagonal  of  a  polyhe- 
dron is  a  straight  line  joining  any  two 
vertices  not  in  the  same  face,  as  AB. 

719.  Defs.     A  polyhedron  of  four 

faces  is  called  a  tetrahedron;  one  of  six  faces,  a  hexahedron; 
one  of  eight  faces,  an  octahedron;  one  of  twelve  faces,  a  do- 
decahedron ;  one  of  twenty  faces,  an  icosahedron ;  etc. 

Ex.  1250.     How  many  diagonals  has  a  tetrahedron  ?  a  hexahedron  ? 

Ex.  1251.  What  is  the  least  number  of  faces  that  a  polyhedron  can 
have  ?  edges  ?  vertices  ? 

Ex.  1252.  How  many  edges  has  a  tetrahedron  ?  a  hexahedron  ?  an 
octahedron  ? 

Ex.  1253.  How  many  vertices  has  a  tetrahedron  ?  a  hexahedron  ? 
an  octahedron  ? 

Ex.  1254.  If  E  represents  the  number  of  edges,  F  the  number  of 
faces,  and  F  the  number  of  vertices  in  each  of  the  polyhedrons  mentioned 
in  Exs.  1252  and  1253,  show  that  in  each  case  E  +  2  =  V  +  F.  This 
result  is  known  as  Euler's  theorem. 

343 


344 


St)LID   GEOMETRY 


Ex.  1255.  Show  that  in  a  tetrahedron  S=  (F—  2)  4  right  angles, 
where  S  is  the  sum  of  the  face  angles  and  V  is  the  number  of  vertices. 

Ex.  1256.  Does  the  formula,  S  =  ( V  -  2)  4  right  angles,  hold  for 
a  hexahedron  ?  an  octahedron  ?  a  dodecahedron  ? 


720.  Def.     A    regular   polyhedron   is   a   polyhedron    all    of 
whose  faces  are  equal  regular  polygons,  and  all  of  whose  polyhe- 
dral angles  are  equal. 

721.  Questions.     How  many  equilateral  triangles  can  meet  to  form 
a  polyhedral  angle  (§  712)  ?     Then  what  is' the  greatest  number  of  regular 
polyhedrons  possible  having  equilateral  triangles  as  faces  ?     What  is  the 
greatest  number  of  regular  polyhedrons  possible  having  squares  as  faces  ? 
having  regular  pentagons  as  faces  ?     Can  a  regular  polyhedron  have  as 
faces  regular  polygons  of  more  than  five  sides  ?   why  ?     What,  then,  is 
the  maximum  number  of  kinds  of  regular  polyhedrons  possible  ? 

722.  From  the  questions  in  §  721,  the  student  has  doubtless 
drawn  the  conclusion  that  not  more  than  five  kinds  of  regular 
polyhedrons  exist.      He  should  convince  himself  that  these 
five  are  possible  by  actually  making  them  from  cardboard  as 
indicated  below : 


Tetrahedron     Hexahedron    Octahedron    Dodecahedron  Icosahednm 


BOOK  VII 


345 


723.  Historical    Note.     The  Pythagoreans  knew  that  there  were 
five  regular  polyhedrons,  but  it  was  Euclid  who  proved  that  there  can  be 
only  jive.    Ilippasus  (arc.  470  B.C.),  who  discovered  the  dodecahedron,  is 
said  to  have  been  drowned  for  announcing  his  discovery,  as  the  Pythago- 
reans were  pledged  to  refer  the  glory  of  any  new  discovery  "  back  to  the 
founder." 

PEISMS  * 

724.  Def.     A  prismatic  surface  is  a  surf  ace .  generated  by  a 
moving  straight  line  that  continually  intersects  a  fixed  broken 
line  and  remains  parallel  to  a  fixed  straight  line  not  coplanar 
with  the  given  broken  line. 


B  C 

Prism  Prismatic  Surface 

725.  Defs.     By  referring  to  §  693,  the  student  may  give  the 
definitions  of  generatrix  and  directrix  of  a  prismatic  surface. 
Point  these  out  in  the  figure. 

726.  Def.     A  prism  is  a  polyhedron  whose  boundary  consists 
of  a  prismatic  surface  and   two  parallel  planes  cutting  the 
generatrix  in  each  of  its  positions. 

727.  Defs.     The  two  parallel  plane  sections  are  the  bases  of 
the  prism,  as  ABODE  and  FGHKL ;  the  faces  forming  the  pris- 
matic surface  are  the  lateral  faces,  as  AG,  BH,  etc.;  the  inter- 
sections o*f  the  lateral  faces  are  the  lateral  edges,  as  AF,  BG,  etc. 

In  this  text  only  prisms  whose  bases  are  convex  polygons 
will  be  considered. 


*  This  treatment  of  prisms  and  pyramids  is  given  because  of  its  similarity  to  the  treat- 
ment of  cylinders  and  cones  given  in  §§  SID-S'2'2  and  t>37-S40. 


346 


SOLID   GEOMETRY 


728.  Def.  A  right  section  of  a  prism  is  a  section  formed  by 
a  plane  which  is  perpendicular  to  a  lateral  edge  of  the  prism 
and  which  cuts  the  lateral  edges  or  the  edges  prolonged. 


Right  Prism  Regular  Prism  Oblique  Prism 

729.  Def.     A  right  prism  is  a  prism  whose  lateral  edges  are 
perpendicular  to  the  bases. 

730.  Def.     A  regular  prism  is  a  right  prism  whose  bases  are 
regular  polygons. 

731.  Def.     An  oblique  prism  is  a  prism  whose  lateral  edges 
are  oblique  to  the  bases. 

732.  Defs.      A  prism  is  triangular,  quadrangular,  etc.,  accord- 
ing as  its  bases  are  triangles,  quadrilaterals,  etc. 

733.  Def.     The  altitude  of  a  prism  is  the  perpendicular  from 
any  point  in  the  plane  of  one  base  to  the  plane  of  the  other  base. 

734.  The  following  are  some  of  the  properties  of  a  prism ; 
the  student  should  prove  the  correctness  of  each  : 

(a)  Any  two  lateral  edges  of  a  prism  are  parallel. 
(6)  The  lateral  edges  of  a  prism  are  equal. 

(c)  Any  lateral  edge  of  a  right  prism  is  equal  to  the  altitude. 

(d)  TJie  lateral  faces  of  a  prism  are  parallelograms. 

(e)  The  lateral  faces  of  a  right  prism  are  rectangles. 
(/)   TJie  bases  of  a  prism  are  equal  polygons. 

(g)  TJie  sections  of  a  prism  made  by  two  parallel  planes  cutting 
all  the  lateral  edges  are  equal  polygons. 

(h)  Every  section  of  a  prism  made  by  a  plane  parallel  to  the 
base  is  equal  to  the  base. 


BOOK  VII 
PROPOSITION  I.     THEOREM 


347 


735.  Two  prisms  are  equal  if  three  faces  including  a 
trihedral  angle  of  one  are  equal  respectively,  and  simi- 
larly placed,  to  three  faces  including  a  trihedral  angle  of 


the  other. 


H 


A  B  A'  B 

Given   prisms  AI  and  A1 11,  face  AJ=  face  A'j',  face  AG  =  face 
'G'J  face  AD  =  face  A'D'. 
To  prove    prism  AI=  prism  A'l'. 


ARGUMENT 

1.  A  BAF,  FAE,  and  BAE  are  equal,  respec- 

tively, to  A  B'A'F',  F'A'E',  and  B'A'E'. 

2.  .-.  trihedral  Z  A  =  trihedral  Z  A1. 

3.  Place  prism  AI  upon  prism  A' I1  so  that 

trihedral  Z  A  shall   be  superposed 
upon  its  equal,  trihedral  Z  A'. 

4.  Faces   AJ,  AG,  and  AD  are  equal,  re- 

spectively, to  faces  A '«/',  A'G',  and  A'D'. 

5.  .  *.  J,  F,  and  G  will  fall  upon  «/',  F',  and 

(?',  respectively. 

6.  CH  and  <?'#'  are  both  ||  BG. 

7.  .-.  Ctfand  C'H'  are  collinear. 

8.  .-.  H  will  fall  upon  .H'. 

9.  Likewise  I  will  fall  upon  /'. 

10.  .*.  prism  AI  =  prism  A'l'.  Q.E.D. 


REASONS 

1.  §  110. 

2.  §  704. 

3.  §  54,  14. 


4.  By  hyp. 

5.  §18. 

6.  §  734,  a. 

7.  §  179. 

8.  §  603,  b. 

9.  By  steps  si m- 

•   ilarto6-8. 

10.  §  18. 


348 


SOLID   GEOMETRY 


736.  Def.     A  truncated  prism  is  the  portion  of  a  prism  in- 
cluded between  the  base  and  a  section  of  the 

prism  made  by  a  plane  oblique  to  the  base, 
but  which  cuts  all  the  edges  of  the  prism. 

737.  Cor.   I.    Two   truncated   prisms 
are  equal  if  three  faces   including  a 
trihedral  angle  of  o^^e   are   equal   re- 
spectively to  three  faces   including   a 
trihedral  angle  of  the  other,  and  the 
faces  are  similarly  placed. 

738.  Cor.  n.     Two  right  prisms  are  equal  if  tf-  ey  have 
equal  bases  and  equal  altitudes. 


Ex.  1257.  Two  triangular  prisms  are  equal  if  their  lateral  faces  are 
equal,  each  to  each. 

Ex.  1258.  Classify  the  polyhedrons  whose  faces  are  :  (a)  four  tri- 
angles ;  (6)  two  triangles  and  three  parallelograms  ;  (c)  two  quadri- 
laterals and  four  parallelograms ;  (c?)  two  quadrilaterals  and  four 
rectangles  ;  (e)  two  squares  and  four  rectangles. 

Ex.  1259.  Find  the  sum  of  the  plane  angles  of  the  dihedral  angles 
whose  edges  are  the  lateral  edges  of  a  triangular  prism  ;  a  quadrangular 
prism.  (HINT.  Draw  a  rt.  section  of  the  prism.) 

Ex.  1260.  Every  section  of  a  prism  made  by  a  plane  parallel  to  a 
lateral  edge- is  a  parallelogram. 

Ex.  1261.  Every  section  of  a  prism  made  by  a  plane  parallel  to  a 
lateral  face  is  a,  parallelogram. 

Ex.  1262.  The  section  of  a  parallelepiped  made  by  a  plane  passing 
through  two  diagonally  opposite  edges  is  a  parallelogram. 


Oblique  Parallelepiped    Right  Parallelepiped     Rectangular  Cube 

Parallelepiped 

739.   Def.     A   parallelepiped   is   a   prism   whose    bases   are 
parallelograms. 


BOOK  VII  349 

740.  Def.     A  right  parallelepiped  is  a  parallelepiped  whose 
lateral  edges  are  perpendicular  to  the  bases. 

741.  Def.     A  rectangular  parallelepiped  is  a  right  parallele- 
piped whose  bases  are  rectangles. 

742.  Def.     A  cube  (i.e.  a  regular  hexahedron)  is  a  rectangu- 
lar parallelepiped  whose  edges  are  all  equal. 

743.  The  following  are  some  of  the  properties  of  a  parallel- 
epiped ;  the  student  should  prove  the  correctness  of  each : 

(a)  All  the  faces  of  a  parallelepiped  are  parallelograms. 

(b)  All  the  faces  of  a  rectangular  parallelopiped  are  rectangles. 

(c)  All  the  faces  of  a  cube  are  squares. 

(d)  Any  two  opposite  faces  of  a  parallelopiped  are  equal  and 
parallel. 

(e)  Any  two  opposite  faces  of  a  parallelopiped  may  be  taken 
as  the  bases. 

Ex.  1263.  Classify  the  polyhedrons  whose  faces  are  :  (a)  six  paral- 
lelograms ;  (&)  six  rectangles  ;  (c)  six  squares  ;  (d)  two  parallelograms 
and  four  rectangles  ;  (e)  two  rectangles  and  four  parallelograms  ;  (/)  two 
squares  and  four  rectangles. 

Ex.  1264.     Find  the  sum  of  all  the  face  angles  of  a  parallelopiped. 

Ex.  1265.     Find  the  diagonal  of  a  cube  whose  edge  is  8  ;  12 ;  e. 

Ex.  1266.  Find  the  diagonal  of  a  rectangular  parallelopiped  whose 
edges  are  6,  8,  and  12  ;  whose  edges  are  a,  &,  and  c. 

Ex.  1267.  The  edge  of  a  cube  :  the  diagonal  of  a  face  :  the  diagonal 
of  the  cube  =  1 :  x  :  y  ;  find  x  and  y. 

Ex.  1268.     Find  the  edge  of  a  cube  whose  diagonal  is  20  \/3  ;  d. 

Ex.  1269.     The  diagonals  of  a  rectangular  parallelopiped  are  equal. 

Ex.  1270.     The  diagonals  of  a  parallelopiped  bisect  each  other. 

Ex.  1271.     The  diagonals  of  a  parallelopiped  meet  in  a  point. 

This  point  is  sometimes  called  the  center  of  the  parallelopiped. 

Ex.  1272.  Any  straight  line  through  the  center  of  a  parallelopiped, 
with  its  extremities  in  the  surface,  is  bisected  at  the  center. 

Ex.  1273.  The  sum  of  the  squares  of  the  four  diagonals  of  a  rectan- 
gular parallelopiped  is  equal  to  the  sum  of  the  squares  of  the  twelve  edges. 

Ex.  1274.     Is  the  statement  in  Ex.  1273  true  for  any  parallelopiped  ? 


350 


SOLID   GEOMET11Y 


D 


PYRAMIDS 

744.  Def.    A  pyramidal  surface  is  a  surface  generated  by  a  mov- 
ing straight  line  that  continually  intersects  a  fixed  broken  line 
and  that  passes  through 

a  fixed  point  not  in  the 
plane  of  the  broken  line. 

745.  Defs.      By   re- 
ferring to  §§  693   and 
694,    give    the    defini- 
tions   of    generatrix, 
directrix,    vertex,    and 
element  of  a  pyramidal 
surface.     Point     these 
out  in  the  figure. 

746.  Def.   A  pyram- 
idal    surface    consists 
of  two  parts  lying  on 
opposite   sides  of    the 
vertex,    called  the  up- 
per and  lower  nappes. 

747.  Def.     A  pyramid  is  a  polyhedron  whose  boundary  con- 
sists of  the  -portion  of  a  pyramidal  surface  extending  from  its 
vertex  to  a  plane  cutting  all 

its  elements,  and  the  section 
formed  by  this  plane. 

748.  Defs.     By  referring 
to  §  727,  the    student   may 
give  the  definitions  of  base, 
lateral  faces,  and  lateral  edges 
of   a  pyramid.     The  vertex 

of  the  pyramidal  surface  is  called  the  vertex  of  the  pyramid, 
as  F.     Point  these  out  in  the  figure. 

In  this  text  only  pyramids  whose  bases  are  convex  polygons 
will  be  considered. 


BOOK   VII 


351 


749.  Defs.     A  pyramid  is  triangular,  quadrangular,  etc.,  ac- 
cording as  its  base  is  a  triangle,  a  quadrilateral,  etc. 

750.  Questions.     How  many  faces  has  a  triangular  pyramid  ?  a  tet- 
rahedron ?     Can  these  terms  be  used  interchangeably  ?     How  many  dif- 
ferent bases  may  a  triangular  pyramid  have  ? 

751.  Def.     The  altitude  of  a  pyramid  is  the  perpendicular 
from  the  vertex  to  the  plane  of  the  base,  as  VO  in  the  figure 
below,  and  in  the  figure  on  preceding  page. 


Regular 
Pyramid 


Truncated  Frustum  of  Frustum  of 

Pyramid     Triangular  Pyramid    Regular  Pyramid 


752.  Def.     A  regular  pyramid  is  a  pyramid  whose  base  is  a 
regular  polygon,  and  whose  vertex  lies  in  the  perpendicular 
erected  to  the  base  at  its  center. 

753.  Def.     A  truncated  pyramid  is  the  portion  of  a  pyramid 
included  between  the  base  and  a  section  of  the  pyramid  made 
by  a  plane  cutting  all  the  edges. 

754.  Def.     A  frustum  of  a  pyramid  is  the  portion  of  a  pyra- 
mid included  between  the  base  and  a  section  of  the  pyramid 
made  by  a  plane  parallel  to  the  base. 

755.  The  following  are  some  of  the  properties  of  a  pyramid; 
the  student  should  prove  the  correctness  of  each : 

(a)  Tlie  lateral  edges  of  a  regular  pyramid  are  equal. 

(b)  TJie  lateral  edges  of  a  frustum  of  a  regular  pyramid  are  equal. 

(c)  The  lateral  faces  of  a  regular  pyramid  are  equal  isosceles 
triangles. 

(d)  TJie  lateral  faces  of  a  frustum  of  a  regular  pyramid  are 
equal  isosceles  trapezoids. 


352 


SOLID   GEOMETRY 


PROPOSITION  II.     THEOREM 

756.   If  a  pyramid  is  cut  by  a  plane  parallel  to  the  base : 
I-    The  edges  and  altitude  are  divided  proportionally. 
II.    The  section  is  a  polygon  similar  to  the  base. 


Given   pyramid  F—  ABODE  and  plane  MN  \\  base  AD  cutting 
the  lateral  edges  in  F,  G,  H,  I,  and  J  and  the  altitude  in  P. 
VA  _  VB  _  VC  _  _VO 

VF~  VG~  VH~  ~  VP 

II.    FGHIJ~ABCDE. 


To  prove : 


I.  ARGUMENT 

1.  Through  F  pass  plane  RS  II  plane  KL. 

2.  Then  plane  RS  II  plane  MN. 

3. 


rA_VB     VB  =  FC      VC__VO    ^ 
"    VF~~  VG'    VG  ~~  F//'    VH~  VP' 


_.__ 
VF~~  VG~~  VH~ 


VO 
VP 


=  — .         Q.E.D. 


REASONS 

1.  §  652. 

2.  §  654. 

3.  §  650. 

4.  §  54,  1. 


II.    The  proof  of  II  is  left  as  an  exercise  for  the  student. 

757.  Cor  I.  Any  section  of  a  pyramid  parallel  to  the 
base  is  to  the  base  as  the  square  of  its  distance  from  the 
vertex  is  to  the  square  of  the  altitude  of  the  pyramid. 

Wr2     "FT?2    ~FP2 
HINT.     Prove  1-,  =  ^  =  —. 


BOOK  VII 


353 


758.  Cor.   II.      If  two  pyramids  having  equal  altitudes 
are  cut  by  planes  parallel  to  their  bases,  and  at  equal  dis- 
tances from  their  vertices, 

the  sections  have  the  same 
ratio  as  the  bases. 

HINT.     Apply  §  757  to  each 
pyramid. 

759.  Cor.  III.      //   two 
pyramids   have  equiva- 
lent bases  and  equal  al- 
titudes, sections  made  by 

planes  parallel  to  the  bases,  and  at  equal  distances  from 
the  vertices,  are  equivalent. 


Ex.  1275.  Is  every  truncated  pyramid  a  frustum  of  a  pyramid?  Is 
every  frustum  of  a  pyramid  a  truncated  pyramid?  What  is  the  lower 
base  of  a  frustum  of  a  pyramid  ?  the  upper  base  ?  the  altitude  ? 

Ex.  1276.    Classify  the  figures  whose  faces  are  as  indicated  below  : 

(a)  one  quadrilateral  and  four  triangles  ; 

(&)  one  square  and  four  equal  isosceles  triangles  ; 

(c)  one  pentagon  and  five  triangles  ; 

(d)  two  pentagons  and  five  trapezoids  ; 

(e)  two  squares  and  four  equal  isosceles  trapezoids  ; 
(/)  two  regular  hexagons  and  six  rectangles. 


Ex.  1277.      In  the  figure  of 
and  VB  =  25,  find  VF  and  VG. 


758,  if  FP=12,  P0  =  8,   VA  =  28, 


Ex.  1278.  The  base  of  a  pyramid,  whose  altitude  is  2  decimeters, 
contains  200  square  centimeters.  Find  the  area  of  a  section  6  centimeters 
from  the  vertex  ;  10  centimeters  from  the  vertex. 

Ex.  1279.  The  altitude  of  a  pyramid  with  square  base  is  16  inches  ; 
the  area  of  a  section  parallel  to  the  base  and  10  inches  from  the  vertex  is 
56-£  square  inches.  Find  the  area  of  the  base. 

Ex.  1280.  The  altitude  of  a  pyramid  is  H.  At  what  distance  from 
the  vertex  must  a  plane  be  passed  parallel  to  the  base  so  that  the  section 
formed  shall  be  :  (a)  one  half  as  large  as  the  base  ?  (6)  one  third  ? 
(c)  one  ?ith  ? 


354 


SOLID   GEOMETRY 


Ex.  1281.  Prove  that  parallel  sections  of  a  pyramid  are  to  each 
other  as  the  squares  of  their  distances  from  the  vertex  of  the  pyramid. 
Do  the  results  obtained  in  Ex.  1280  fulfill  this  condition  ? 

Ex.  1282.  Each  side  of  the  base  of  a  regular  hexagonal  pyramid 
is  6  ;  the  altitude  is  15.  How  far  from  the  vertex  must  a  plane  be  passed 
parallel  to  the  base  to  form  a  section  whose  area  is  12  v£? 

Ex.  1283.  The  areas  of  the  bases  of  a  frustum  of  a  pyramid  are  288 
square  feet  and  450  square  feet ;  the  altitude  of  the  frustum  is  3  feet. 
Find  the  altitude  of  the  pyramid  of  which  the  given  figure  is  a  frustum. 

Ex.  1284.  The  bases  of  a  frustum  of  a  regular  pyramid  are  equi- 
lateral triangles  whose  sides  are  10  inches  and  18  inches,  respectively  ; 
the  altitude  of  the  frustum  is  8  inches.  Find  the  alti- 
tude of  the  pyramid  of  which  the  given  figure  is  a 
frustum. 

Ex.  1285.  The  sum  of  the  lateral  faces  of  any 
pyramid  is  greater  than  the  base. 

HINT.  In  the  figure,  let  VE  be  the  altitude  of  face 
VAD  and  VO  the  altitude  of  the  pyramid.  Which  is 
the  greater,  VE  or  OE  ? 


MENSURATION  OF  THE  PRISM  AND  PYRAMID 

AREAS 


760.  Def.     The  lateral  area  of  a  prism, 
turn  of  a  pyramid  is  the  sum  of  the  areas 

761.  In  the  mensuration  of  the  prism 
lowing  notation  will  be  used : 

a,  &,  c  =  dimensions  of  a  rectangu- 
lar parallelepiped. 

B  =  area  of  base  in  general  or  of 
lower  base  of  a  frustum. 

b  =  area  of  upper  base  of  a 
frustum. 

E  =  lateral  edge,  or  element, 
or  ecjge  of  a  tetrahedron 
in  general. 

H  =  altitude  of  a  solid. 

h  =  altitude  of  a  surface. 

L  =  slant  height. 


0  = 
P  = 


P  = 


T  = 
V  = 


a  pyramid,  or  a  frus- 
of  its  lateral  faces. 

and  pyramid  the  fol- 


vertex  of  a  pyramid. 

perimeter  of  right  sec- 
tion or  of  the  lower 
base  of  a  frustum. 

perimeter  of  upper  base 
of  a  frustum. 

lateral  area. 

total  area. 

volume  in  general. 

volumes  of  smaller  sol- 
ids into  which  a  larger 
solid  is  divided. 


BOOK  VII 


355 


PROPOSITION  III.     THEOREM 

762.   The  lateral  area  of  a  prism  is  equal  to  the  product 
of  the  perimeter  of  a  right  section  and  a  lateral  edge. 


M 


Given   prism  AK  with  MQ  a  rt.  section,  E  a  lateral  edge,  S  the 
lateral  area,  and  P  the  perimeter  of  rt.  section  MQ. 
To  prove    S  =  P  •  E. 


ARGUMENT 

1.  Rt.  section  MQ  _L  AI,  CJ,  etc. 

2.  .-.  MN±.AI;  NQA-CJ;  etc. 

3.  .-.  MN  is  the  altitude  of  O^J";  NQ  is 


4. 


the  altitude  of  O  CK\  etc. 
area  of  O  AJ  =  MN  •  AI  =  MN  -  E-, 
area  of  O  CK  =  NQ  •  CJ  =  NQ  •  E\ 

etc. 


S  =  P  •  E. 


REASONS 

1.  §728. 

2.  §619. 

3.  §228. 

4.  §481. 


5.  §54,2. 

6.  §309. 


763.  Cor.     The  lateral  area  of  a  right  prism  is  equal  to 
the  product  of  the  perimeter  of  its  base  and  its  altitude. 

HINT.     Thus,  if  P  =  perimeter  of  base  and  //  =  altitude,  S  =  P  •  H. 

764.  Def.     The  slant  height  of  a  regular  pyramid  is  the  alti- 
tude of  any  one  of  its  triangular  faces. 

765.  Def.     The  slant  height  of  a  frustum  of  a  regular  pyra- 
mid is  the  altitude  of  any  one  of  its  trapezoidal  faces. 


356 


SOLID   GEOMETRY 


PROPOSITION  IV.     THEOREM 

766.  The  lateral  area  of  a  regular  pi/ramid  is  equal  to 
one  half  the  product  of  the  perimeter  of  its  base  and 
its  slant  height. 


C        D 

Given   regular  pyramid  0-ACD  •••  with  the  perimeter  of  its 
base  denoted  by  P,  its  slant  height  by  L,  and  its  lateral  area  by  8. 
To  prove    S  =  1  P  •  L. 


ARGUMENT 

1.  Area  of  AAOC=  \  AC  -  OR  =  \  AC 

area  of  A  COD  —  1  CD  •  L ;  etc. 

2.  /.  AAOC  +  ACOD+  ... 


3. 


= 


L. 


Q.E.D. 


REASONS 

1.  §  485. 

2.  §  54,  2. 

3.  §  309. 


W 


767.  Cor.  The  lateral  area  of 
a  frustum  of  a  regular  pyramid 
is  equal  to  one  half  the  product 
of  the  sum  of  the  perimeters  of 
its  bases  and  its  slant  height. 

HINT.     Prove  8  =  |(P  +  p)L. 

Ex.  1286.  Find  the  lateral  area  and 
the  total  area  of  a  regular  pyramid  each 
side  of  whose  square  base  is  24  inches,  and 
whose  altitude  is  16  inches. 

Ex.  1287.  The  sides  of  the  bases  of  a  frustum  of  a  regular  octagonal 
pyramid  are  15  centimeters  and  24  centimeters,  respectively,  and  the 
slant  height  is  30  centimeters.  Find  the  number  of  square  decimeters  in 
the  lateral  area  of  the  frustum. 


D 


BOOK   VII  357 

Ex.  1288.  Find  the  lateral  area  of  a  prism  whose  right  section  is  a 
quadrilateral  with  sides  5,  7,  9,  and  13  inches,  and  whose  lateral  edge  is 
15  inches. 

Ex.  1289.  Find  the  lateral  area  of  a  right  prism  whose  altitude  is  16 
inches  and  whose  base  is  a  triangle  with  sides  8,  11,  and  13  inches. 

Ex.  1290.  The  perimeter  of  a  right  section  of  a  prism  is  45  deci- 
meters ;  its  altitude  is  10  V3  decimeters  ;  and  a  lateral  edge  makes  with 
the  base  an  angle  of  60°.  Find  the  lateral  area. 

Ex.  1291.  Find  the  altitude  of  a  regular  prism,  one  side  of  whose 
triangular  basjp  is  5  inches  and  whose  lateral  area  is  195  square  inches. 

Ex.  1292.  Find  the  total  area  of  a  regular  hexagonal  prism  whose 
altitude  is  20  inches  and  one  side  of  whose  base  is  10  inches. 

Ex.  1293.     Find  the  total  area  of  a  cube  whose  diagonal  is  8V3. 

Ex.  1294.  Find  the  edge  of  a  cube  if  its  total  area  is  294  square 
centimeters  ;  if  its  total  area  is  T. 

Ex.  1295.  Find  the  total  area  of  a  regular  tetrahedron  whose  edge 
is  6  inches. 

Ex.  1296.  Find  the  lateral  area  and  total  area  of  a  regular  tetra- 
hedron whose  slant  height  is  8  inches. 

Ex.  1297.  Find  the  lateral  area  and  total  area  of  a  regular  hexagonal 
pyramid,  aside  of  whose  base  is  6  inches  and  whose  altitude  is  10  inches. 

Ex.  1298.  Find  the  total  area  of  a  rectangular  parallelepiped  whose 
edges  are  6,  8,  and  12  ;  whose  edges  are  a,  ft,  and  c. 

Ex.  1299.  Find  the  total  area  of  a  right  parallelepiped,  one  side  of 
whose  square  base  is  4  inches,  and  whose  altitude  is  6  inches. 

Ex.  1300.  The  balcony  of  a  theater  is  supported  by  four  columns 
whose  bases  are  regular  hexagons.  Find  the  cost,  at  2  cents  a  square 
foot,  of  painting  the  columns  if  they  are  20  feet  high  and  the  apothems  of 
the  bases  are  10  inches. 

Ex.  1301.  In  a  frustum  of  a  regular  triangular  pyramid,  the  sides 
of  the  bases  are  8  and  4  inches,  respectively,  and  the  altitude  is  10  inches. 
Find  the  slant  height  and  a  lateral  edge. 

Ex.  1302.  In  a  frustum  of  a  regular  hexangular  pyramid,  the  sides 
of  the  bases  are  12  and  8,  respectively,  and  the  altitude  is  16.  Find  the 
lateral  area. 

Ex.  1303.  In  a  regular  triangular  pyramid,  the  side  of  the  base  is 
8  inches,  and  the  altitude  is  12  inches  ;  a  lateral  face  makes  with  the  base 
an  angle  of  60°.  Find  the  lateral  area. 


358 


SOLID   GEOMETRY 


VOLUMES 

768.  Note.    The  student  should  compare  carefully  §§  769-776  with 
the  corresponding  discussion  of  the  rectangle,  §§  466-473. 

769.  A  solid  may  be  measured  by  finding  how  many  times 
it  contains  a  solid  unit.     The  solid  unit  most  frequently  chosen 
is  a  cube  whose  edge  is  of  unit  length.     If  the  unit  length  is 
an  inch,  the  solid  unit  is  a  cube  whose  edge  is  an  inch.     Such 
a  unit  is  called  a  cubic  inch.     If  the  unit  length  is  a  foot,  the 
solid  unit  is  a  cube  whose  edge  is  a  foot,  and  the  unit  is  called 
a  cubic  foot. 


FIG.  1.     Rectangular  Parallelepiped  AD  =  60  U. 

770.  Def.      The  result   of  the    measurement  is  a  number, 
which  is  called  the  measure-number,  or  numerical  measure,  or 
volume  of  the  solid. 

771.  Thus,  if  the  unit  cube  U  is  contained  in  the  rectangular 


Fie.  2.     Rectangular  Parallelepiped  AD  =  24  U+. 


BOOK  VII 


359 


parallelepiped  AD  (Fig.  1)  60  times,  then  the  measure-number  or 
volume  of  rectangular  parallelepiped  AD,  in  terms  of  u,  is  60. 

If  the  given  unit  cube  is  not  contained  in  the  given  rectangu- 
lar parallelepiped  an  integral  number  of  times  without  a  re- 
mainder (Fig.  2),  then  by  taking  a  cube  that  is  an  aliquot  part 
of  u,  as  one  eighth  of  U,  and  applying  it  as  a  measure  to  the 
rectangular  parallelepiped  (Fig.  3),  a  number  will  be  obtained 


A  2" 

FIG.  3.     Rectangular  Parallelepiped  AD  =  ^-  U+  =  39|  Z7+. 

which,  divided  by  8,*  will  give  another  (and  usually  closer) 
approximate  volume  of  the  given  rectangular  parallelepiped. 
By  proceeding  in  this  way  (Fig.  4),  closer  and  closer  approxi- 
mations to  the  true  volume  may  be  obtained. 


FIG.  4.    Rectangular  Parallelepiped  AD  =  *% %&  U+  =  41  &  77+. 
*  It  takes  eight  of  the  small  cubes  to  make  the  unit  cube  itself. 


360  SOLID  GEOMETRY 

772.  If  the  edges  of  the  given  rectangular  parallelepiped 
and  the  edge  of  the  unit  cube  are  commensurable,  a  cube  may 
be  found  which  is  an  aliquot  part  of  U,  and  which  will  be  con- 
tained in  the  rectangular  parallelopiped  an  integral  number  of 
times. 

773.  If  the  edges  of  the  given  rectangular  parallelopiped 
and  the  edge  of  the  unit  cube  are  incommensurable,  then  closer 
and  closer  approximations  to  the  volume  may  be  obtained,  but 
no  cube  which  is  an  aliquot  part  of  U  will  be  also  an  aliquot 
part  of  the  rectangular  parallelopiped  (by  definition  of  incom- 
mensurable magnitudes). 

There  is,  however,  a  definite  limit  which  is  approached  more 
and  more  closely  by  the  approximations  obtained  by  using 
smaller  and  smaller  subdivisions  of  the  unit  cube,  as  these 
subdivisions  approach  zero  as  a  limit. 

774.  Def.    The  volume  of  a  rectangular  parallelopiped  which 
is  incommensurable  with   the  chosen  unit  cube  is  the  limit 
which    successive   approximate    volumes   of    the    rectangular 
parallelopiped  approach  as  the  subdivisions  of  the  unit  cube 
approach  zero  as  a  limit. 

For  brevity  the  expression  the  volume  of  a  solid,  or  simply 
the  solid,  is  used  to  mean  the  volume  of  the  solid  with  respect 
to  a  chosen  unit. 

775.  Def.    The  ratio  of  any  two  solids  is  the  ratio  of  their 
measure-numbers,  or  volumes  (based  on  the  same  unit). 

776.  Def.    Two  solids  are  equivalent  if   their  volumes  are 
equal. 

777.  Historical   Note.      The    determination    of    the   volumes    of 
polyhedrons  is  found  in  a  document  as  ancient  as  the  Rhind  papyrus, 
which  is  thought  to  be  a  copy  of  a  manuscript  dating  back  possibly  as  far 
as  3400  B.C.   (See  §  474.)     In  this  manuscript  Ahmes  calculates  the  con- 
tents of  an  Egyptian  barn  by  means  of  the  formula,  V  —  a-  &•  (c  +^  c), 
where  a,  b,  and  c  are  supposed  to  be  linear  dimensions  of  the  barn.     But 
unfortunately  the  exact  shape  of  these  barns  is  unknown,  so  that  the 
accuracy  of  the  formula  cannot  be  tested. 


BOOK   VII 


361 


PROPOSITION  V.     THEOREM 

778.   The  volume  of  a  rectangular  parallelopiped  is 
equal  to  the  product  of  its  three  dimensions. 


K 


Given  rectangular  parallelopiped  AD,  with  dimensions  AC, 
AF,  and  A  G ;  and  U  the  chosen  unit  of  volume,  whose  edge  is  u. 

To  prove    the  volume  of  AD  =  AC  -  AF  •  AG. 

I.    If  AC,  AF,  and  AG  are  each  commensurable  with  u. 

(a)  Suppose  that  u  is  contained  in  AC,  AF,  and  AG  each  an 
integral  number  of  times. 


ARGUMENT 

1.  Lay  off  u  upon  AC,  AF,  and  AG.     Sup- 

pose that  u  is  contained  in  ACr  times, 
in  AF  s  times,  and  in  AG  t  times. 

2.  At  the  points  of  division  on  AC,  AF,  and 

AG  draw  planes  _L  AC,  AF,  and  AG. 

3.  Then  AD  is  divided  into  unit  cubes. 

4.  There  are  r  of  these  unit  cubes  in  a 

row  along  AC,  s  of  these  rows  in 
parallelopiped  AK,  and  t  such  paral- 
lelepipeds in  parallelopiped  AD. 

5.  .-.  the  volume  of  AD  =  r  •  s  •  t. 

6.  But  r,  s,  and  t  are  the  measure-numbers 

of  AC,  AF,  and  AG,  respectively,  re- 
ferred to  the  linear  unit  u. 

7.  .-.  the  volume  of  AD  =  AC-AF-  AG.  Q.E.D. 


REASONS 


1.    §  335. 


2.  §  627. 

3.  §  769. 

4.  Arg.  1. 


5.  §  770. 

6.  Arg.  1. 


7.    §  309. 


362 


SOLID   GEOMETRY 


(7>)  Suppose  that  u  is  not  a  measure  of  AC,  AF,  and  AG,  re- 
spectively, but  that  some  aliquot  part  of  u  is  such  a  measure. 
The  proof  is  left  as  an  exercise  for  the  student. 


II.   If  AC,  AF,  and  AG  are  each  incommensurable  with  u. 


ARGUMENT 

1.  Let  m  be  a  measure  of  u.    Apply  m  as 

a  measure  to  AC,  AF,  and  AG,  respec- 
tively, as  many  times  as  possible. 
There  will  be  remainders,  as  MC, 
NF,  and  QG,  each  less  than  m. 

2.  Through    M    draw    plane    MK  J_  AC, 

through  N  draw  plane  NK  A.  AF,  and 
through  Q  draw  plane  QK  _L  AG. 

3.  Now  ^3f,  -4JV,  and  ^Q  are  each  com- 

mensurable with  the  measure  m,  and 
hence  with  u,  the  linear  unit. 

4.  .*.  the  volume  of  rectangular  parallele- 

piped AK  =  AM  •  AN  -  AQ. 

5.  Now  take  a  smaller  measure  of  u.     No 

matter  how  small  a  measure  of  u  is 
taken,  when  it  is  applied  as  a  measure 
to  AC,  AF,  and  ^(?,  the  remainders, 
MC,  NF,  and  QG,  will  be  smaller  than 
tho  Tw.'isure  taken. 


REASONS 
1.    §339. 


2.  §627. 

3.  §337. 

4.  §778,1. 

5.  §335. 


BOOK   VII 


363 


ARGUMENT 

6.  .-.  the  difference  between  AM  and  AC, 

the  difference  between  AN  and  AF, 
and  the  difference  between  AQ  and 
AG,  may  each  be  made  to  become 
and  remain  less  than  any  previously 
assigned  segment,  however  small. 

7.  .-.  AM  approaches  AC  as  a  limit,  AN 

approaches  AF  as  a  limit,  and  AQ 
approaches  AG  as  a  limit. 

8.  .-.  AM-  AN-  AQ  approaches  AC'  AF  •  AG 

as  a  limit. 

9.  Again,  the  difference  between  rectan- 

gular parallelopiped  AK  and  rec- 
tangular parallelepiped  AD  may  be 
made  to  become  and  remain  less  than 
any  previously  assigned  volume, 
however  small. 

10.  .'.  the  volume  of  rectangular  parallelo- 

piped AK  approaches  the  volume  of 
rectangular  parallelopiped  AD  as  a 
limit. 

11.  But  the  volume  of  AK  is  always  equal 

to  AM-  AN  -  AQ. 

12.  .-.  the  volume  of  AD=AC-  AF-  AG.  Q.E.D. 


11. 
12. 


REASONS 
Arg.  5. 


7.    §349. 


§593. 
Arg.  5. 


10.    §349. 


Arg.  4. 
§355. 


III.  If  AC  is  commensurable  with  u  but  AF  and  AG  are  in- 
commensurable with  u. 

IV.  If  AC  and  AF  are  commensurable  with  u  but  AG  is  in- 
commensurable with  u. 

The  proofs  of  III  and  IV  are  left  as  exercises  for  the  student. 

779.   Cor  I.     The  volume  of  a  cube  is  equal  to  the  cube 
of  its  edge. 

HINT.     Compare  with  §  478. 


364  SOLID   GEOMETRY 

780.  Cor.  II.     Any     two    rectangular    parallelepipeds 
are   to   each   other  as   the  products   of  their   three  di- 
mensions.    (HINT.     Compare  with  §  479. ) 

781.  Note.     By  the  product  of  a  surface  and  a  line  is  meant  the 
product  of  the  measure-numbers  of  the  surface  and  the  line. 

782.  Cor.  III.    The  volume  of  a  rectangular  parallelo- 
piped  is  equal  to  the  product  of  its  base  and  its  altitude. 

783.  Cor.  IV.    Any    two    rectangular    parallelepipeds 
are   to  each  other  as  the  products   of  their  bases    and 
their  altitudes.     (HINT.    Compare  with  §  479.) 

784.  Cor.  V.  (a)  Two  rectangular  parallelepipeds  having 
equivalent  bases  are  to  each  other  as  their  altitudes;  (&) 
two  rectangular  parallelepipeds  having  equal  altitudes 
are  to  each  other  as  their  bases.    (HINT.    Compare  with  §  480.) 

785.  Cor.  VI.     (a}  Two  rectangular  parallelepipeds  hav- 
ing two  dimensions  in  common  are  to  each  other  as  their 
third  dimensions,  and  (b)  two  rectangular  parallelepi- 
peds having  one  dimension  in  common  are  to  each  other 
as  the  products  of  their  other  two  dimensions. 

786.  Questions.     What  is  it  in  Book  IV  that  corresponds  to  volume 
in  Book  VII  ?  to  rectangular  parallelepiped  ?    State  the  theorem  and  corol- 
laries in  Book  IV  that  correspond  to  §§  778,  779,  782,  783,  and  784.     Will 
the  proofs  given  there,  with  the  corresponding  changes  in  terms,  apply 
here  ?     Compare  the  entire  discussion  of  §§  466-480  with  §§  769-785. 


Ex.  1304.     Find  the  volume  of  a  cube  whose  diagonal  is  5\/3  ;  d. 

Ex.  1305.  The  volume  of  a  rectangular  parallelepiped  is  V ;  each 
side  of  the  square  base  is  one  third  the  altitude  of  the  parallelepiped.  Find 
the  side  of  the  base.  Find  the  side  of  the  base  if  V=  192  cubic  feet. 

Ex.  1306.  The  dimensions  of  two  rectangular  parallelepipeds  are  6, 
8,  10  and  5,  12,  16,  respectively.  Find  the  ratio  of  their  volumes. 

Ex.  1307.    The  total  area  of  a  cube  is  300  square  inches ;  find  its  volume. 

Ex.  1308.  The  volume  of  a  certain  cube  is  V\  find  the  volume  of  a 
cube  whose  edge  is  twice  that  of  the  given  cube. 

Ex.  1309.  The  edge  of  a  cube  is  a  ;  find  the  edge  of  a  cube  twice  as 
large  ;  i.e.  containing  twice  the  volume  of  the  given  cube. 


BOOK  VII 


365 


PLATO 


787.  Historical  Note.  Plato  (429-348  B.C.)  was  one  of  the  first  to 
discover  a  solution  to  that  famous  problem  of  antiquity,  the  duplication 
of  a  cube,  i.e.  the  finding  of  the 
edge  of  a  cube  whose  volume  is 
double  that  of  a  given  cube. 

There  are  two  legends  as  to 
the  origin  of  the  problem.  The 
one  is  that  an  old  tragic  poet  rep- 
resented King  Minos  as  wishing 
to  erect  a  tomb  for  his  son  Glau- 
cus.  The  king  being  dissatis- 
fied with  the  dimensions  (100 
feet  each  way)  proposed  by  his 
architect,  exclaimed  :  "  The  in- 
closure  is  too  small  for  a  royal 
tomb  ;  double  it,  but  fail  not  in 
the  cubical  form." 

The  other  legend  asserts  that 
the  Athenians,  who  were  suf- 
fering from  a  plague  of  typhoid  fever,  consulted  the  oracle  at  Delos  as  to 
how  to  stop  the  plague.  Apollo  replied  that  the  Delians  would  have 
to  double  the  size  of  his  altar,  which  was  in  the  form  of  a  cube.  A  new 
altar  was  constructed  having  its  edge  twice  as  long  as  that  of  the  old  one. 
The  pestilence  became  worse  than  before,  whereupon  the  Delians  ap- 
pealed to  Plato.  It  is  therefore  known  as  the  Delian  problem. 

Plato  was  born  in  Athens,  and  for  eight  years  was  a  pupil  of  Socrates. 
Plato  possessed  considerable  wealth,  and  after  the  death  of  Socrates  in 
399  B.C.  he  spent  some  years  in  traveling  and  in  the  study  of  mathe- 
matics.. It  was  during  this  time  that  he  became  acquainted  with  the 
members  of  the  Pythagorean  School,  especially  with  Archytas,  who  was 
then  its  head.  No  doubt  it  was  his  association  with  these  people  that  gave 
him  his  passion  for  mathematics.  About  380  B.C.  he  returned  to  his  native 
city,  where  he  established  a  school.  Over  the  entrance  to  his  school  was 
this  inscription:  "Let  none  ignorant  of  geometry  enter  my  door." 
Later  an  applicant  who  knew  no  geometry  was  actually  turned  away 
with  the  statement :  "  Depart,  for  thou  hast  not  the  grip  of  philosophy." 

Plato  is  noted  as  a  teacher,  rather  than  an  original  discoverer,  and  his 
contributions  to  geometry  are  improvements  in  its  method  rather  than 
additions  to  its  matter.  He  valued  geometry  mainly  as  a  "means  of  ed- 
ucation in  right  seeing  and  thinking  and  in  the  conception  of  imaginary 
processes."  It  is  stated  on  good  authority  that  "Plato  was  almost  as 
important  as  Pythagoras  to  the  advance  of  Greek  geometry." 


366  SOLID   GEOMETRY 

PROPOSITION  VI.     THEOREM 

788.  An  oblique  prism  is  equivalent  to  a  right  prism, 
whose  base  is  a  right  section  of  the  oblique  prism,  and 
whose  altitude  is  equal  to  a  lateral  edge  of  the  oblique 
prism. 


4, 

B          C 

Given  oblique  prism  AD' ;  also  rt.  prism  KNf  with  base  KN 
a  rt.  section  of  AD',  and  with  KKf,  LLf,  etc.,  lateral  edges  of 
KN',  equal  to  AA',  BB1,  etc.,  lateral  edges  of  AD\ 

To  prove  oblique  prism  AD1  =c=  rt.  prism  KN1. 

OUTLINE  OF  PROOF 

1.  In  truncated  prisms  AN  and  A'N',  prove  the  A  of   face 
BK  equal,  respectively,  to  the  A  of  face  B'K'. 

2.  Prove  the  sides  of  face  BK  equal,  respectively,  to  the 
sides  of  face  B'K'. 

3.  .-.  face  BK=f&ce  B'K'. 

4.  Similarly  face  53f=face  B'M',  and  face  BE  =  face  B'E'. 

5.  .-.  truncated  prisrn  AN  =  truncated  prism  A'N'  (§  737). 

6.  But  truncated  prism  A'N  =  truncated  prism  A'N. 

7.  .-.  oblique  prism  AD'  =c=rt.  prism  KN'.  Q.E.D. 

789.  Question.  Is  there  a  theorem  in  Book  IV  that  corresponds 
to  Prop.  VI  ?  If  not,  formulate  one  and  see  if  you  can  prove  it  true. 


BOOK   VII 


367 


PROPOSITION  VII.     THEOREM 

790.    The  volume  .of  any  parallelepiped   is   equal   to 
the  product  of  its  base  and  its  altitude. 


Given  parallelepiped  I  with  its  volume  denoted  by  F,  its  base 
by  Ji,  and  its  altitude  by  H. 
To  prove  V  =  B  •  H. 


ARGUMENT 

1.  Prolong  edge  4C  and  all  edges  of  I II  AC. 

2.  On  the  prolongation  of  AC  take  DF  = 

AC,  and  through  D  and  ^  pass  planes 
J_  ^,  forming  rt.  parallelepiped  17. 

3.  Then  I  =c=  77. 

4.  Prolong  edge  -FJT  and  all  edges  of  //  II  FK. 

5.  On  the  prolongation  of  FK  take  MN  = 

FK,  and  through  J/  and  N  pass  planes 
J_  FN,  forming  rectangular  parallele- 
piped ///. 

6.  Then    II  ^  III. 

7.  .:  1=0=1/7. 

8.  Again,  J?  =0=  B'  =  B". 

9.  Also  Hj  the  altitude  of  7,  =  the  altitude 

of  ///. 

10.  But  the  volume  of  777=  B"  •  H. 

11.  .-.  v—  B  •  H.  Q.E.D. 


REASONS 

1.  §54,16. 

2.  §  627. 


3.  §  788. 

4.  §  54,  16. 

5.  §  627. 


6.  §  788. 

7.  §54,1. 

8.  §  482. 

9.  §  663. 


10. 
11. 


§782. 
§  309. 


368  SOLID   GEOMETRY 

791.  Cor.  I.    Parallelopipeds  having  equivalent   bases 
and  equal  altitudes  are  equivalent. 

792.  Cor.  II.    Any    two    parallelepipeds    are    to    each 
other  as  the  products  of  their  bases  and  their  altitudes. 

793.  Cor.  III.    (a)  Two  parallelepipeds  having  equiva- 
lent   bases   are   to  each  other  as  their  altitudes,   and 
(b}  two  parallelepipeds   having  equal  altitudes  are  to 
each  other  as  their  bases. 

794.  Questions.     What  expression  in  Book  IV  corresponds  to  vol- 
ume of  a  parallelepiped  9    Quote  the  theorem  and  corollaries  in  Book  IV 
that  correspond  to  §§  790-793.     Will  the  proofs  given  there,  with  the 
corresponding  changes,  apply  here  ? 


Ex.  1310.  Prove  Prop.  VI  by  subtracting  the  equal  truncated  prisms 
of  Arg.  5  from  the  entire  figure. 

Ex.  1311.  The  base  of  a  parallelepiped  is  a  parallelogram  two  adja- 
cent sides  of  which  are  8  and  15,  respectively,  and  they  include  an  angle 
of  30°.  If  the  altitude  of  the  parallelepiped  is  10,  find  its  volume. 

Ex.  1312.  Four  parallelepipeds  have  equivalent  bases  and  equal  lat- 
eral edges.  In  the  first  the  lateral  edge  makes  with  the  base  an  angle  of 
30°  ;  in  the  second  an  angle  of  45°  ;  in  the  third  an  angle  of  60°  ;  and  in 
the  fourth  an  angle  of  90°.  Find  the  ratio  of  the  volumes  of  the  four 
parallelepipeds. 

Ex.  1313.  Find  the  edge  of  a  cube  equivalent  to  a  rectangular  par- 
allelepiped whose  edges  are  6,  10,  and  15  ;  whose  edges  are  a,  &,  and  c. 

Ex.  1314.  Find  the  diagonal  of  a  cube  whose  volume  is  512  cubic 
inches  ;  a  cubic  inches. 

Ex.  1315.  The  edge  of  a  cube  is  a.  Find  the  area  of  a  section  made 
by  a  plane  through  two  diagonally  opposite  edges. 

Ex.  1316.  How  many  cubic  feet  of  cement  will  be  needed  to  make  a 
box,  including  lid,  if  the  inside  dimensions  of  the  box  are  2  feet  6  inches, 
3  feet,  and  4  feet  6  inches,  if  the  cement  is  3  inches  thick  ? 

HINT.  In  a  problem  of  this  kind,  always  find  the  volume  of  the  whole 
solid,  and  the  volume  of  the  inside  solid,  then  subtract. 

Ex.  1317.  The  volume  of  a  rectangular  parallelepiped  is  2430  cubic 
inches,  and  its  edges  are  in  the  ratio  of  3,  6,  and  6.  Find  its  edges. 

Ex.  1318.  In  a  certain  cube  the  area  of  the  surface  and  the  volume 
have  the  same  numerical  value.  Find  the  volume  of  the  cube. 


BOOK  VII 


369 


PROPOSITION  VIII.     THEOREM 

795.    The  plane  passed  through  two  diagonally  opposite 
edges  of  a  parallelopiped  divides  it  into  two  equivalent 


triangular  prisms 


Given  plane  AG  passed  through  edges  AE  and  CG  of  paral- 
lelopiped BH  dividing  it  into  the  two  triangular  prisms 
ABC-F  and  CDA-H. 

To  prove    prism  ABC-F  o  prism  CDA-H. 

ARGUMENT  ONLY 

1.  Let  MNOP  be  a  rt.  section  of  parallelopiped  BH,  cutting 
the  plane  AG  in  line  MO. 

2.  Face  AF  II  face  DG  and  face  AH\\  face  BG. 

3.  .  •.  MN  II  PO  and  MP  II  NO. 

4.  .  • .  MNOP  is  a  O. 

5.  .'.A  MNO  =  A  0PM. 

6.  Now  triangular  prism  ABC-F  =c=  a  rt.  prism  whose  base  is 
A  MNO,  a  rt.  section  of  prism  ABC-F,  and  whose  altitude  is  AE, 
a  lateral  edge  of  prism  ABC-F. 

7.  Likewise  triangular  prism  CDA-Hoa,  rt.  prism  whose  base 
is  A  0PM  and  whose  altitude  is  AE. 

8.  But  two  such  prisms  are  equivalent. 

9.  .-.  prism  ABC-F  =c=  prism  CDA-H.  Q.E.D. 

796.  Questions.  Is  there  a  theorem  in  Book  I  that  corresponds  to 
Prop.  VIII  ?  If  so,  state  it.  Could  an  oblique  prism  exist  such  that  a 
right  section,  as  MNOP,  might  intersect  either  base  ?  If  so,  draw  a 
figure  to  illustrate. 


370 


SOLID   GEOMETRY 


PROPOSITION  IX.     THEOREM 

797.    The  volume  of  a  triangular  prism/  is  equal  to 
the  product  of  its  base  and  its  altitude. 


A  C 

Given  triangular  prism  A  CD-X  with  its  volume  denoted  by 
F,  its  base  by  B,  and  its  altitude  by  H. 

To  prove    F  =  B  •  H. 

The  proof  is  left  as  an  exercise  for  the  student. 

798.  Questions.  What  proposition  in  Book  IV  corresponds  to  Prop. 
IX  above  ?  Can  you  apply  the  proof  there  given  ?  What  is  the  name  of 
the  figure  CZ  in  §  797  ?  What  is  its  volume  ?  What  part  of  CZ  is 
AGD-X  (§  795)  ?  

Ex.  1319.  The  volume  of  a  triangular  prism  is  equal  to  one  half  the 
product  of  any  lateral  face  and  the  perpendicular  from  any  point  in  the 
opposite  edge  to  that  face. 

HINT.  The  triangular  prism  is  one  half  of  a  certain  parallelepiped 
(§  795). 

Ex.  1320.  The  base  of  a  coal  bin  which  is  8  feet  deep  is  a  triangle 
with  sides  10  feet,  15  feet,  and  20  feet,  respectively.  How  many  tons  of 
coal  will  the  bin  hold  considering  35  cubic  feet  of  coal  to  a  ton  ? 

Ex.  1321.  One  face  of  a  triangular  prism  contains  45  square  inches  ; 
the  perpendicular  to  this  face  from  a  point  in  the  opposite  edge  is  6  inches. 
Find  the  volume  of  the  prism. 

Ex.  1322.  During  a,  rainfall  of  \  inch,  how  many  barrels  of  water 
will  fall  upon  a  ten-acre  field,  counting  7£  gallons  to  a  cubic  foot  and  31^ 
gallons  to  a  barrel  ? 

Ex.  1323.  The  inside  dimensions  of  an  open  tank  before  lining  are 
6  feet,  2  feet  0  inches,  and  2  feet,  respectively,  the  latter  being  the  height. 
Find  the  number  of  pounds  of  zinc  required  to  line  the  tank  with  a  coat- 
ing \  inch  thick,  a  cubic  foot  of  zinc  weighing  6860  ounces. 


BOOK  vn 

THEOREM 


371 


PROPOSITION  X. 

799.   The  volume  of  any  prism  is  equal  to  the  product 
of  its  base  and  its  altitude, 


c> 

Given   prism  AM  with  its  volume  denoted  by  F,  its  base  by 
B,  and  its  altitude  by  H. 
To  prove    F  =  B  •  H. 

ARGUMENT 

1.  From  any  vertex  of  the  lower  base,  as 

Ay  draw  diagonals  AD,  AF,  etc. 

2.  Through  edge  AI  and  these  diagonals 

pass  planes  AK,  AM,  etc. 

3.  Prism  AM  is  thus  divided  into  triangu- 

lar prisms. 

4.  Denote  the  volume  and  base  of  trian- 

gular prism  ACD-J  by  vl  and  bi',  of 
ADF-K  by  v2  and  62 ;  etc.     Then 


v  = 


V  =  B  -  H. 


etc. 


Q.E.D. 


REASONS 

1.  §54,15. 

2.  §  612. 

3.  §  732. 

4.  §  797. 


5.  §  54,  2. 

6.  §  309. 


800.  Cor.  I.     Prisms    having    equivalent     bases     and 
equnl  altitudes  are  equivalent. 

801.  Cor.  n.     Any    two   prisms   are   to   each  other   as 
the  products  of  their  bases  and  their  altitudes. 

802.  Cor.  m.     (a)  Two  prisms  having  equivalent  bases 
are   to   each  other   as   their   altitudes;     (&)  tivo  prisms 
having  equal  altitudes  are  to  each  other  as  their  bases. 


372 


SOLID   GEOMETRY 


PROPOSITION  XI.     THEOREM 

803.   Two  triangular  pyramids  having  equivalent  bases 
and  equal  altitudes  are  equivalent. 

o  R 


Given  triangular  pyramids  0-ACD  and  O'-A'C'D'  with  base 
ACD  =c=  base  A'C'D',  with  altitudes  each  equal  to  QR,  and  with 
volumes  denoted  by  Fand  F',  respectively. 

To  prove    V  —  V1. 


ARGUMENT 

1.  V—  F',  F<  F',  or  F>  F'. 

2.  Suppose  F  <  F',  so  that  v1  —  V  =  ~k,  a 

constant.  For  convenience,  place 
the  two  pyramids  so  that  their  bases 
are  in  the  same  plane,  MN. 

3.  Divide  the  common  altitude  QR  into  71 

equal  parts,  as  QX,  XY,  etc.,  and 
through  the  several  points  of  division 
pass  planes  II  plane  MN. 

4.  Then    section    FGU  =e=  section    F'G'U', 

section  JKW  =c=  section  J'K'  w',  etc. 

5.  On  FGU,  JKW,  etc.,  as  upper  bases,  con- 

struct prisms  with  edges  II  DO  and 
with  altitudes  =  QX.  Denote  these 
prisms  by  II,  III,  etc. 
0.  On  A'c'D',  F'G'u',  etc.,  as  lower  bases, 
construct  prisms  with  edges  II  D'o' 
and  with  altitudes  =  QX.  Denote 
these  prisms  by  I',  II';  etc. 


REASONS 

1.  161,  a. 

2.  §  54,  14. 


3.  §653. 

4.  §759. 

5.  §726. 

6.  §726. 


BOOK  VII 


373 


9. 
10. 
11. 


ARGUMENT 

7.  Then  prism  II  =0=  prism  II',  prism  III 

=c=  prism  III',  etc. 

8.  Now  denote  the  sum  of  the  volumes  of 

prisms  II,  III,  etc.,  by  /S;  the  sum 
of  the  volumes  of  prisms  I',  II',  III', 
etc.,  by  S1 ;  and  the  volume  of  prism 
I'  by  v'.  Then  S'  -  S  =  v'. 

But  V1  <  S'  and  S  <  F. 

.-.  V*  +  S  <  F  +  S'. 

.'.  V1  —  V  <  S'  —  S  ;  i.e.  V*  —  V<v'. 
12.  By  making  the  divisions  of  the  altitude 
QR  smaller  and  smaller,  prism  I', 
and  hence  v'  may  be  made  less  than 
any  previously  assigned  volume, 
however  small. 

F'  -  -  F,  which  is  <  v',  may  be  made 
less   than   any  previously  assigned 
volume,  however  small, 
the  supposition  that  v'  —  V=~k,  a 
constant,  is  false ;  i.e.  V  is  not  <  F'. 

15.  Similarly  it  may  be  proved  that  V9  is 

not  <  F. 

16.  .-.  F=  Ff.  Q.E.D. 


13. 


14. 


REASONS 

7.  §800. 

8.  §54,3. 


9.  §54,12. 

10.  §54,9. 

11.  §54,5. 

12.  802,  a. 


13.    §  54,  10. 


14.  Arg.  13. 

15.  By  steps  sim- 

ilar to  2-14. 

16.  §  161,  b. 


Ex.  1324.  The  volume  of  an  oblique  prism  is  equal  to  the  product  of 
its  right  section  and  a  lateral  edge. 

HINT.     Apply  §  788. 

Ex.  1325.  The  volume  of  a  regular  prism  is  equal  to  the  product  of 
its  lateral  area  and  one  half  the  apothem  of  its  base. 

HINT.     See  Ex.  1319. 

Ex.  1326.  The  base  of  a  prism  is  a  rhombus  having  one  side  29  inches 
and  one  diagonal  42  inches.  If  the  altitude  of  the  prism  is  25  inches, 
find  its  volume. 

Ex.  1327.  In  a  certain  cube  the  area  of  the  surface  and  the  com- 
bined lengths  of  its  edges  have  the  same  numerical  value.  Find  the 
volume  of  the  cube. 


374 


SOLID   GEOMETRY 


PROPOSITION  XII.     TJIEOREM 

804.   The  volume  of  a  triangular  pyramid  is  equal  to 
one  third  the  product  of  its  base  and  its  altitude. 


Given   triangular  pyramid  0-ACD  with  its  volume  denoted 
by  F,  its  base  by  £,  and  its  altitude  by  H. 
To  prove  F  =  |  B  •  H. 

ARGUMENT  ONLY 

1.  Construct  prism  AG  with  base  ACD  and  lateral  edge  CO. 

2.  The    prism    is    then    composed  of  triangular  pyramid 
0-ACD  and  quadrangular  pyramid  O-ADGF. 

3.  Through  OD  and  OF  pass  a  plane,  intersecting  ADGF  in 
DP  and  dividing  quadrangular  pyramid  O-ADGF  into  two  tri- 
angular pyramids  0-ADF  and  0-DGF. 

4.  ADGF  is  &  [3-,    .-.  A  ADF  =  ADGF. 

5.  /.  0-ADF  =0=  0-DGF. 

6.  But  in  triangular  pyramid  0-DGF,  OGF  may  be  taken  as 
base  and  D  as  vertex ;  then  0-DGF  =  D-OGF  =c=  0-ACD. 

1.    But  0-ACD  +  0-ADF  -f-  0-DGF  =c=  prism  AG. 

8.  .-.  3  times  the  volume  of  O-ACD=the  volume  of  prism  AG. 

9.  /.  F=  1  the  volume  of  prism  AG. 

10.   But  prism  AG  =  B>H;  .-.  V=  ±B- H.  Q.K.D. 


Ex.  1328.  Find  the  volume  of  a  regular  tetra- 
hedron whose  edge  is  0. 

HINT.  0,  the  foot  of  the  _L  fromFto  the  plane 
of  base  ABC,  is  the  center  of  A  ABC  (§  752). 
Hence  OA  =  f  of  the  altitude  of  A  ABC,  and  a  _L 
from  O  to  any  edge  of  the  base,  as  OD  =  \  of  OA. 

Ex.  1329.  Find  tlie  volume  of  a  regular  tetra- 
hedron with  slant  height  2  V3  ;  with  altitude  a. 


BOOK  VII  375 

PROPOSITION  XIII.     THEOREM 

805.   The  volume  of  any  pyramid  is  equal  to  one  third 
the  product  of  its  base  and  its  altitude, 

O 


A 

C D 

Given   pyramid  0-ACDFG  with  its  volume  denoted  by  F,  its 
base  by  n,  and  its  altitude,  OQ,  by  H. 
To  prove     F=  ^  B  •  H. 

The  proof  is  left  as  an  exercise  for  the  student. 
HINT.    See  proof  of  Prop.  X. 

806.  Cor.  I.    Pyramids   having    equivalent    bases  and 
equal  altitudes  are  equivalent. 

807.  Cor.  II.    Any  two  pyramids  are  to  each  other  as  the 
products  of  their  bases  and  their  altitudes. 

808.  Cor.  III.  (a)  Two  pyramids  having  equivalent  bases 
are  to  each  other  as  their  altitudes,  and  (b}  two  pyramids 
having  equal  altitudes  are  to  each  other  as  their  bases. 

Ex.  1330.  In  the  figure  of  §  805,  if  the  base  =  250  square  inches,  OO 
=  18  inches,  and  the  inclination  of  OC  to  the  base  is  60°,  find  the  volume. 

Ex.  1331.  A  pyramid  and  a  prism  have  equivalent  bases  and  equal 
altitudes  ;  find  the  ratio  of  their  volumes. 


809.  Historical  Note.  The  proof  of  the  proposition  that  "every 
pyramid  is  the  third  part  of  a  prism  on  the  same  base  and  with  the  same 
altitude  "  is  attributed  to  Eudoxus  (408-355  B.C.),  a  great  mathematician 
of  the  Athenian  School.  In  a  noted  work  written  by  Archimedes  (287- 
212  B.C.),  called  Sphere  and  Cylinder,  there  is  also  found  an  expression 
for  the  surface  and  volume  of  a  pyramid.  (For  a  further  account  of 
Archimedes,  see  §§  542,  896,  and  973.)  Later  a  solution  of  this  problem 
was  given  by  Brahmagupta,  a  noted  Hindoo  writer  born  about  598  A. DO 


376 


SOLID   GEOMETRY 


PROPOSITION  XIV.     THEOREM 


810.  Two  triangular  pyramids,  having  a  trihedral 
angle  of  one  equal  to  a  trihedral  angle  of  the  other,  are  to 
each  other  as  the  products  of  the  edges  including  the 
equal  trihedral  angles.  D 


Given  triangular  pyramids  0-ACD  and  Q-FGM  with  tri- 
hedral Z  0  ==  trihedral  Z  Q,  and  with  volumes  denoted  by  F 
and  F',  respectively. 

F        OA  -  OO-  OD 
To  prove     — ,  = . 

F'       QF-QG-QM 


ARGUMENT 

REASONS 

1. 

Place   pyramid  Q-FGM  so  that  trihedral 

1. 

§  54,  14. 

Z  Q  shall  coincide  with  trihedral  Z  0. 

Eepresent  pyramid  Q-FGM  in   its  new 

position  by  O-F'G'M'. 

2. 

From  J9  and  J/'  draw  DJ  and  J/'.K'  _L  plane  OA  C. 

2. 

§  639. 

3 

Then  F  —  A  OAC  '  DJ    —  A  °^a    -0*7 

Q 

S  807 

»J. 

F'  ~~  A  OF'G1  •  3/'#      A  OF'G'   M'K 

o» 

3  ou  i  . 

4 

But  A  ^       °^  '  °C  - 

4 

S  4.Q8 

A  O^'G?'     OF'  -  OG' 

T:. 

S  itt/o. 

5. 

Again  let  the  plane  determined  by  DJ  and 

5. 

§§613,616. 

M'K  intersect  plane  OA  C  in  line  OKJ. 

6. 

Then  rt.  A  DJO  ~  rt.  A  M*KO. 

6. 

§  422. 

7. 

DJ  _  OD 

'  M'K~~  OM1' 

7. 

§  424,  2. 

ft 

V       OA-OC      OD        OA-OC-OD 

'           ~~~                      •            ~~                                    Q  E  D 

Q 

S  30Q 

o. 

F'     OF'-OG'  OM'     QF.QG-QM' 

O. 

J5    <./\/  »7  • 

811.  Def.  Two  polyhedrons  are  similar  if  they  have  the 
same  number  of  faces  similar  each  to  each  and  similarly  placed, 
and  have  their  corresponding  polyhedral  angles  equal. 


BOOK   VII 


377 


PROPOSITION  XV.     THEOREM 

812.   The  volumes  of  two  similar  tetrahedrons  are  to 
each  other  as  t}ie  cubes  of  any  two  homologous  edges. 


M 


Given  similar  tetrahedrons  0—ACD  and  Q-FGM  with  volumes 
denoted  by  F  and  F*,  and  with  OA  and  QF  two  homologous 
edes. 


_  Y  iy_i3. 

To  prove      —  =  — — . 
V      QF3 

ARGUMENT 

1.    Trihedral  Z  O  =  trihedral  Z  Q. 
2  F  _  O4  -  OC  -  OP  _  OA      OC      OP 

'  V'  ~  QF  -  QG  -  QM  ~~  QF     QG      QJ/  ' 

3.  But^  =  ^  =  ^. 

QF      QG       QM 

V       OA      OA      OA       ~OA3 

4.  .'.   —  = . .  =  — — .  Q.E.D. 

V'       QF      QF     QF       QF3 

813.  Question.     Compare  §§  810  and  812  with  §§  498  and  503.     Are 
the  same  general  methods  used  in  the  two  sets  of  theorems  ? 

814.  Note.     The  proposition,  "  two  similar  convex  polyhedrons  are 
to  each  other  as  the  cubes  of  any  two  homologous  edges,"  will  be  assumed 
at  this  point,  and  will  be  applied  in  some  of  the  exercises  that  follow.     For 
a  complete  discussion  of  this  principle  see  Appendix,  §§  1022-1029. 


REASONS 

1.  §  811. 

2.  §  810. 

3.  §  424,  2. 

4.  §  309. 


Ex.  1332.  The  edges  of  two  regular  tetrahedrons  are  6  centimeters 
and  8  centimeters,  respectively.  Find  the  ratio  of  their  volumes. 

Ex.  1333.  The  volumes  of  two  similar  polyhedrons  are  343  cubic 
inches  and  512  cubic  inches,  respectively:  (a)  an  edge  of  the  first  figure  is 
14  inches,  find  the  homologous  edge  of  the  second  ;  (&)  the  total  area  of 
the  first  figure  is  280  square  inches,  find  the  total  area  of  the  second. 


378 


SOLID   GEOMETRY 
PROPOSITION  XVI.     THEOREM 


815.  The  volume  of  a  frustum  of  any  pyramid  is 
equal  to  one  third  the  product  of  its  altitude  and  the 
sum  of  its  lower  base,  its  upper  base,  and  the  mean  pro- 
portional between  its  two  bases. 


C  D 

Given  frustum  AM,  of  pyramid  0-AF,  with  its  volume  de- 
noted by  F,  its  lower  base  by  B,  its  upper  base  by  b,  and  its 
altitude  by  H. 

To  prove    V=  $H(B 


+  V  'B  •  b). 


ARGUMENT 


1.  Frustum    AM  =  pyramid    0-AF    minus 

pyramid  0-RM. 

2.  Let  H'  denote  the  altitude  of  0-RM. 
Then        F^=  1  B(H  +  H')  —  1  b  •  H' 

=  i  HB  4-  i  H'(B  —  b). 
It  now  remains  to  find  the  value  of  H'. 
b  H'2 


3. 
4. 

5.    Whence  H1  = 
G.  .-.  v  = 


B       (H+  H1)'' 


II' 


i.e.  F=  J  //(« 


-  b  ; 


lf  -6).  Q.K.I). 


REASONS 

1.  §  54,  11. 

2.  §  805. 

3.  §  757. 

4.  §  54,  13. 

5.  Solving  for  if'. 
G.  §  309. 


BOOK  VII  379 

PROPOSITION  XVII.     THEOREM 

816.  Jl  truncated  triangular  prism  is  equivalent  to 
three  triangular  pyramids  whose  bases  are  the  base  of 
the  frustum  and  whose  vertices  are  the  three  vertices 
of  the  inclined  section. 

K 


Given   truncated  triangular  prism  ACD-FGK. 

To  prove    ACD-FGK  =c=  F-ACD  +  G-ACD  +  K-ACD. 


ARGUMENT 

1.  Through  A,  D,  F  and  K,  D,  F  pass  planes 

dividing  frustum  ACD-FGK  into 
three  triangular  pyramids  F-ACD, 
F-ADK,  and  F-DGK.  Since  F-ACD  is 
one  of  the  required  pyramids,  it  re- 
mains to  prove  F-ADK  =0=  K-ACD  and 
F-DGK  =0=  G-ACD. 

2.  CF  II  plane  AG. 

3.  .-.  the   altitude   of   pyramid   F-ADK  = 

the  altitude  of  pyramid  C-ADK. 

4.  .-.  F-ADK  =o  C-ADK. 

5.  But  in   C-ADK,  A  CD  may  be  taken  as 

base  and  K  as  vertex. 

6.  .'.  F-ADK  =0=  K-ACD. 

7.  Likewise     F-DGK  =c=  C-DGK  =  K-CDGj 

and  K-CDG  =0=  A-CDG  =  G-ACD.\ 

8.  .'.  F-DGK  =c=  G-ACD. 

9.  .-.  .4<7Z>  —  TOS-  =0=  F-ACD   +    O-^CD   + 

K-ACD.  Q.E.D. 


REASONS 
1.    §611. 


§646. 
§664. 


4.  §806. 

5.  §§749,750. 


§309. 

By  steps  sim- 
ilar to  3-7. 
§  54, 1. 
§309. 


380  SOLID   G-EOMETRY 

817.  Cor.  I.     The  volume  of  a  truncated  right  triangu- 
lar prism  is  equal  to  one  third  the  product  of  its  base 
and  the  sum  of  its  lateral  edges. 

818.  Cor.  H.      The    volume   of  any 
truncated  triangular  prism  is  equal 
to  one  third  the  product  of  a  right  sec- 
tion and  the  sum  of  its  lateral  edges. 

HINT.  Rt.  section  ACD  divides  truncated 
triangular  prism  QE  into  two  truncated  right 
triangular  prisms. 

Ex.  1334.  The  base  of  a  truncated  right  triangular  prism  has  for  its 
sides  13,  14,  and  15  inches  ;  its  lateral  edges  are  8,  11,  and  13  inches.  Find 
its  volume. 

Ex.  1335.  In  the  formula  of  §  815 :  (1)  put  6  =  0  and  compare 
result  with  formula  of  §  805 ;  (2)  put  b  =  B  and  compare  result  with 
formula  of  §  799. 

Ex.  1336.  A  frustum  of  a  square  pyramid  has  an  altitude  of  13 
inches  ;  the  edges  of  the  bases  are  2|  inches  and  4  inches,  respectively. 
Find  the  volume. 

Ex.  1337.  The  edges  of  the  bases  of  a  frustum  of  a  square  pyramid 
are  3  inches  and  5  inches,  respectively,  and  the  volume  of  the  frustum  is 
204£  cubic  inches.  Find  the  altitude  of  the  frustum. 

Ex.  1338.  The  base  of  a  pyramid  contains  144  square  inches,  and  its 
altitude  is  10  inches.  A  section  of  the  pyramid  parallel  to  the  base 
divides  the  altitude  into  two  equal  parts.  Find :  (a)  the  area  of  the 
section;  (6)  the  volume  of  the  frustum  formed. 

Ex.  1339.  A  section  of  a  pyramid  parallel  to  the  base  cuts  off  a  pyra- 
mid similar  to  the  given  pyramid. 

Ex.  1340.  The  total  areas  of  two  similar  tetrahedrons  are  to  each 
other  as  the  squares  of  any  two  homologous  edges. 

Ex.  1341.  The  altitude  of  a  pyramid  is  6  inches.  A  plane  parallel  to 
the  base  cuts  the  pyramid  into  two  equivalent  parts.  Find  the  altitude  of 
the  frustum  thus  formed. 

Ex.  1342.  Two  wheat  bins  are  similar  in  shape  ;  the  one  holds  1000 
bushels,  and  the  other  800  bushels.  If  the  first  is  15  feet  deep,  how  deep 
is  the  second  ? 

Ex  1343.  A  plane  is  passed  parallel  to  the  base  of  a  pyramid  cut- 
ting the  altitude  into  two  equal  parts.  Find  :  (a)  the  ratio  of  the  section 
to  the  base  ;  (6)  the  ratio  of  the  pyramid  cut  off  to  the  whole  pyramid. 


BOOK  VII  381 

MISCELLANEOUS  EXERCISES 

Ex.  1344.  Find  the  locus  of  all  points  equidistant  from  the  three 
edges  of  a  trihedral  angle. 

Ex.  1345.  Find  the  locus  of  all  points  equidistant  from  the  three 
faces  of  a  trihedral  angle. 

Ex.  1346.  («)  Find  the  ratio  of  the  volumes  and  the  ratio  of  the 
total  areas  of  two  similar  tetrahedrons  whose  homologous  edges  are  in 
the  ratio  of  2  to  5.  (6)  Find  the  ratio  of  their  homologous  edges  and 
the  ratio  of  their  total  areas  if  their  volumes  are  in  the  ratio  of  1  to  27. 

Ex.  1347.  (a)  Construct  three  or  more  equivalent  pyramids  on  the 
same  base.  (6)  Find  the  locus  of  the  vertices  of  all  pyramids  equivalent 
to  a  given  pyramid  and  standing  on  the  same  base. 

HINT.     Compare  with  Exs.  821  and  822. 

Ex.  1348.  The  altitude  of  a  pyramid  is  12  inches.  Its  base  is  a 
regular  hexagon  whose  side  is  5  inches.  Find  the  area  of  a  section  paral- 
lel to  the  base  and  4  inches  from  the  base  ;  4  inches  from  the  vertex. 

Ex.  1349.     A  farmer  has  a  corn  crib  20  feet  long,  a  cross  section  of 
which  is  represented  in  the  figure,  the  numbers  denoting  feet.     If  the  crib 
is  entirely  filled  with  corn  in  the  ear,  how  many  bushels 
of  corn  will  it  contain,  counting  2  bushels  of  corn  in  the 
ear  for   1   bushel   of  shelled  corn.     (Use  the   approxi- 
mation, 1  bushel  =  l\  cubic  feet.     For  the  exact  volume 
of  a  bushel,  see  Ex.  1439.) 

Ex.  1350.     A  wheat  elevator  in  the  form  of  a  frus- 
tum of  a  square  pyramid  is  30  feet  high  ;  the  edges  of  its 
bases  are   12  feet  and  6  feet,  respectively.     How  many 
bushels  of  wheat  will  it  hold?     (Use  the  approximation  given  in  Ex. 
1349. ) 

Ex.  1351.  A  frustum  of  a  regular  square  pyramid  has  an  altitude  of 
12  inches,  and  the  edges  of  its  bases  are  4  inches  and  10  inches,  respec- 
tively. Find  the  volume  of  the  pyramid  of  which  the  frustum  is  a  part. 

Ex.  1352.  In  a  frustum  of  a  regular  quadrangular  pyramid,  the  sides 
of  the  bases  are  10  and  6,  respectively,  and  the  slant  height  is  14.  Find 
the  volume. 

Ex.  1353.  Find  the  lateral  area  of  a  regular  triangular  pyramid 
whose  altitude  is  8  inches,  and  each  side  of  whose  base  is  6  inches. 

Ex.  1354.  The  edge  of  a  cube  is  a.  Find  the  edge  of  a  cube  3  times 
as  large  ;  n  times  as  large. 


382 


SOLID   GEOMETRY 


36 


50 


Ex.  1355.  A  berry  box  supposed  to  contain  a  quart  of  berries  is  in 
the  form  of  a  frustum  of  a  pyramid  5  inches  square  at  the  top,  4|  inches 
square  at  the  bottom,  and  2£  inches  deep.  The  United  States  dry  quart 
contains  67.2  cubic  inches.  Does  the  box  contain  more  or  less  than  a 
quart  ? 

Ex.  1356.  The  space  left  in  a  basement  for  a  coal  bin  is  a  rectangle 
8  x  10  feet.  How  deep  must  the  bin  be  made  to  hold  10  tons  of  coal  ? 

Ex.  1357.  The  figure  represents  a  barn,  the  numbers  denoting  the 
dimensions  in  feet.  Find  the  number  of  cubic  feet  in  the  barn. 

Ex.  1358.  Let  AB,  BC,  and 
BD,  the  dimensions  of  the  barn  in 
Ex.  1357,  be  denoted  by  a,  ft,  and  c, 
respectively.  Substitute  the  values 
of  a,  6,  and  c  in  Ahmes'  formula 
given  in  §  777.  Compare  your  result 
with  the  result  obtained  in  Ex.  1357. 

Would  Ahmes'  formula  have  been 

correct  if  the  Egyptian  barns  had  been  similar  in  shape  to  the  barn  in  Ex. 
1357? 

Ex.  1359.  How  much  will  it  cost  to  paint  the  barn  in  Ex.  1357  at 
1  cent  per  square  foot  for  lateral  surfaces  and  2  cents  per  square  foot 
for  the  roof  ?  ^ 

Ex.  1360.  The  barn  in  Ex.  1357  has  a  stone  foundation  18  inches 
wide  and  3  feet  deep.  Find  the  number  of  cubic  feet  of  masonry  if  the 
outer  surfaces  of  the  walls  are  in  the  same  planes  as  the  sides  of  the  barn. 

Ex.  1361.  The  volume  of  a  regular  tetrahedron  is  *f-V2.  Find  its 
edge,  slant  height,  and  altitude. 

Ex.  1362.     The  edge  of  a  regular  octahedron  is  a.     Prove  that  the 

volume  equals—  V2. 
3 

Ex.  1363.  The  planes  determined  by  the  diagonals  of  a  parallelo- 
piped  divide  the  parallelepiped  into  six  equivalent  pyramids. 

Ex.  1364.  A  dam  across  a  stream  is  40  feet  long,  12  feet  high,  7  feet 
wide  at  the  bottom,  and  4  feet  wide  at  the  top.  How  many  cubic  feet  of 
material  are  there  in  the  dam  ?  how  many  loads,  counting  1  cubic  yard  to 
a  load  ?  Give  the  name  of  the  geometrical  solid  represented  by  the  dam. 

Ex.  1365.  Given  8  the  lateral  area,  and  H  the  altitude,  of  a  regular 
square  pyramid,  find  the  volume. 

Ex.  1366.  Find  the  volume  F,  of  a  regular  square  pyramid,  if  its 
total  surface  is  71,  and  one  edge  of  the  base  is  a. 


BOOK  VIII 


CYLINDERS  AND  CONES 

CYLINDERS 

819.  Def.  A  cylindrical  surface  is  a  surface  generated  by  a 
moving  straight  line  that  continually  intersects  a  fixed  curve 
and  remains  parallel  to  a  fixed  straight  line  not  coplanar  with 
the  given  curve. 


N 


Cylindrical  Surface 


FIG.  2.     Cylinder 


820.  Defs.      By  referring  to  §§  693  and  694,  the  student 
may  give  the  definitions  of  generatrix,  directrix,  and  element  of 
a  cylindrical  surface.     Point  these  out  in  the  figure. 

The  student  should  note  that  by  changing  the  directrix  from 
a  broken  line  to  a  curved  line,  a  prismatic  surface  becomes  a 
cylindrical  surface. 

821.  Def.     A  cylinder  is  a  solid  closed  figure  whose  boun- 
dary consists  of  a  cylindrical  surface  and  two  parallel  planes 
cutting  the  generatrix  in  each  of  its  positions,  as  DC. 

383 


384 


SOLID   GEOMETRY 


822.  Defs.     The  two  parallel  plane  sections  are  called  the 
bases  of  the  cylinder,  as  AC  and  DF  (Fig.  4)  ;   the  portion  of 
the  cylindrical  surface  between  the  bases  is  the  lateral  surface 
of  the  cylinder ;  and  the  portion  of  an  element  of  the  cylin- 
drical surface    included   between  the  bases  is  an  element  of 
the  cylinder,  as  MN. 

823.  Def .     A  right  cylinder  is  a  cylinder  whose  elements  are 
perpendicular  to  the  bases. 


FIG.  3.    Right  Cylinder 


FIG.  4.    Oblique  Cylinder 


824.  Def.    An  oblique  cylinder  is  a  cylinder  whose  elements 
are  not  perpendicular  to  the  bases. 

825.  Def.     The  altitude  of  a  cylinder  is  the  perpendicular 
from  any  point  in  the  plane  of  one  base  to  the  plane  of  the 
other  base,  as  HK  in  Figs.  3  and  4. 

826.  The  following  are  some  of  the  properties  of  a  cylinder ; 
the  student  should  prove  the  correctness  of  each : 

(a)  Any  two  elements  of  a  cylinder  are  parallel   and  equal 
(§§  618  and  634). 

(b)  Any  element  of  a  right  cylinder  is  equal  to  its  altitude. 

(c)  A  line  drawn  through  any  point  in  the  lateral  surface  of  a 
cylinder  parallel  to  an  element,  and  limited  by  the  bases,  is  itself  an 
element  (§§  822  and  179). 


BOOK   VIII  385 

PROPOSITION  I.     THEOREM 
827.   The  bases  of  a  cylinder  are  equal. 


Given  cylinder  AD  with  bases  AB  and  CD. 
To  prove  base  AB  =  base  CD. 

ARGUMENT  ONLY 

1.  Through  any  three  points  in  the  perimeter  of  base  AB, 
as  E,  F,  and  G,  draw  elements  EH,  FK,  and  GL. 

2.  Draw  EF,  FG,  GE,  HK,  KL,  and  LH. 

3.  EH  is  II  and  =  FK\  .-.  EK  is  a  O. 

4.  .:  EF=  HK;  likewise  FG  =  KL  and  GE  =LH. 

5.  .'.A  EFG  =  A  JZKL. 

6.  .*.  base  AB  may  be  placed  upon  base  CD  so  that  E,  F,  and 
(?  will  fall  upon  H,  K,  and  L,  respectively. 

7.  But  E,  F,  and  #  are  any  three  points  in  the  perimeter  of 
base  AB ;  i.e.  every  point  in  the  perimeter  of  base  AB  will  fall 
upon  a  corresponding  point  in  the  perimeter  of  base  CD. 

8.  Likewise  it  can  be  shown  that  every  point  in  the  perim- 
eter of  base  CD  will  fall  upon  a  corresponding  point  in  the 
perimeter  of  base  AB. 

9.  .*.  base  AB  may  be  made  to  coincide  with  base  CD. 

10.   .•.  base  AB  =  base  CD.  Q.E.D. 

828.  Cor.  I.     The  sections  of  a  cylinder  made  by  two 
parallel  planes  cutting  all  the  elements  are  equal. 

829.  Cor.  n.     Every  section  of  a  cylinder  made  by  a 
plane  parallel  to  its  base  is  equal  to  the  base. 


386 


SOLID   GEOMETRY 


Ex.  1367.  Every  section  of  a  cylinder  made  by  a  plane  parallel  to 
its  base  is  a  circle,  if  the  base  is  a  circle. 

Ex.  1368.  If  a  line  joins  the  centers  of  the  bases  of  a  cylinder,  this 
line  passes  through  the  center  of  every  section  of  the  cylinder  parallel  to 
the  bases,  if  the  bases  are  circles. 


830.   Def.     A  right  section  of  a  cylinder  is  a  section  formed 
by  a  plane  perpendicular  to  an  element,  as  section  EF. 


FIG.  1.    Cylinder  with 
Circular  Base  AB 


Circular  Cylinder 


FIG.  3.    Eight  Cir- 
cular Cylinder 


831.  Def.     A  circular  cylinder  is  a  cylinder  in  which  a  right 
section  is  a  circle ;  thus,  in  Fig.  2,  if  rt.  section  EF  is  a  O, 
cylinder  AD  is  a  circular  cylinder. 

832.  Def.     A  right  circular  cylinder  is  a  right  cylinder  whose 
base  is  a  circle  (Fig.  3). 

833.  Questions.     In  Fig.  1,  is  rt.  section  EF  a  O  ?    In  Fig.  2,  is 
base  AB  a  O  ?     In  Fig.  3,  would  a  rt.  section  be  a  O  ? 

834.  Note.     The  theorems  and  exercises  on  the  cylinder  that  follow 
will  be  limited  to  cases  in  which  the  bases  of  the  cylinders  are  circles. 
When  the  term  cylinder  is  used,  therefore,  it  must  be  understood  to  mean 
a  cylinder  with  circular  bases.     See  also  §  846. 


Ex.  1369.  Find  the  locus  of  all  points  at  a  distance  of  6  inches  from 
a  straight  line  2  feet  long. 

Ex.  1370.  Find  the  locus  of  all  points :  (a)  2  inches  from  the  lateral 
surface  of  a  right  circular  cylinder  whose  altitude  is  12  inches  and  the 
radius  of  whose  base  is  5  inches  ;  (6)2  inches  from  the  entire  surface. 

Ex.  1371.  A  log  is  20  feet  long  and  30  inches  in  diameter  at  the 
smaller  end.  Find  the  dimensions  of  the  largest  piece  of  square  timber, 
the  same  size  at  each  end,  that  can  be  cut  from  the  log. 


BOOK  VIII 
PROPOSITION  II.     THEOREM 


387 


835.   Every  section  of  a  cylinder  made  by  a  plane  pass- 
ing through  an  element  is  a  parallelogram.    (See  §  834.) 


Given  cylinder  AB  with  base  AK,  and  CDEF  a  section  ma'de 
by  a  plane  through  element  CF  and  some  point,  as  D,  not  in 
CF,  but  in  the  circumference  of  the  base. 

To  prove  CDEF  a  O. 


ARGUMENT 

1.  Through  D  draw  a  line  in  plane  DF  II  CF. 

2.  Then  the  line  so  drawn  is  an  element; 

i.e.  it  lies  in  the  cylindrical  surface. 

3.  But  this  line  lies  also  in  plane  DF. 

4.  /.it  is   the   intersection  of   plane   DF 

with  the  cylindrical  surface,  and  coin- 
cides with  DE. 

5.  .'.  DE  is  a  str.  line  and  is  II  and  =  CF. 

6.  Also  CD  and  EF  are  str.  lines. 

7.  .'.  CDEF  is  a  O.  Q.E.D. 


KEASONS 

§179. 

§  826,  c. 

Arg.  1. 
§  614. 


§  826,  a. 
§  616. 
§  240. 


836.  Cor.  Every  section  of  a  right  circular  cylinder 
made  by  a  plane  passing  through  an  element  is  a  rec- 
tangle.   

Ex.  1372.  In  the  figure  of  Prop.  II,  the  radius  of  the  base  is  4  inches, 
element  CF  is  12  inches,  CD  is  1  inch  from  the  center  of  the  base,  and  CF 
makes  with  CD  an  angle  of  60°.  Find  the  area  of  section  CDEF. 

Ex.  1373.  Every  section  of  a  cylinder,  parallel  to  an  element,  is  a 
parallelogram.  How  is  the  base  of  this  cylinder  restricted  ?  (See  §  834.) 


388 


SOLID   GEOMETRY 


Conical  Surface 


CONES 

837.  Def.     A   conical   surface  is  a  surface  generated  by  a 
moving  straight  line  that  continually  intersects  a  fixed  curve 
and  that  passes  through  a  fixed 

point  not  in  the  plane  of  the 
curve. 

838.  Defs.     By  referring  to 
§§  693,  694,  and  746,  the  stu- 
dent  may  give  the  definitions 
of  generatrix,   directrix,  vertex, 
element,   and  upper    and   lower 
nappes    of    a    conical    surface. 
Point  these  out  in   the  figure. 

The  student  should  observe 
that  by  changing  the  directrix 
from  a  broken  line  to  a  curved 
line,  a  pyramidal  surface  becomes  a  conical  surface. 

839.  Def.     A  cone  is  a  solid  closed  figure 
whose  boundary  consists  of  the  .portion  of  a 
conical  surface  extending  from  its  vertex  to 
a  plane  cutting  all   its   elements,  and  the 
section  formed  by  this  plane. 

840.  Defs.     By  referring  to  §§  748  and 
822,  the   student  may  give  'the  definitions 
of  vertex,  base,  lateral  surface,  and  element 
of  a  cone.     Point  these  out  in  the  figure. 

841.  Def.     A  circular  cone  is  a  cone  containing  a  circular 
section  such  that  a  line  joining  the  vertex  of  the  cone  to  the 
center  of  the  section  is  perpendicular  to  the  section. 

Thus  in  Fig.  4,  if  section  AB  of  cone  V-CD  is  a  O  with  center 
0,  such  that  VO  is  _L  the  section,  cone  V-CD  is  a  circular  cone. 

842.  Def.     The  altitude  of  a  cone  is  the  perpendicular  from 
its  vertex  to  the  plane  of  its  base,  as  VC  in  Fig.  3  and   VO  in 
Fig.  5, 


Fio.  2.    Cone 


BOOK   VIII 


389 


843.   Defs.     In  a  cone  with  a  circular  base,  if  the  line  join- 
ing its  vertex  to  the  center  of  its  base  is  perpendicular  to  the 


FIG.  3.    Cone  with  Circular  Base 


FIG.  4.    Circular  Cone 
V 


FIG.  5.    Right  Circu- 
lar Cone 


plane  of  the  base,  the  cone  is  a  right  cir- 
cular cone  (Fig.  5). 

If  such  a  line  is  not  perpendicular  to  the 
plane  of  the  base,  the  cone  is  called  an 
oblique  cone  (Fig.  3). 

844.  Def.     The  axis  of  a  right  circular 
cone  is  the  line  joining  its  vertex  to  the 
center  of  its  base,  as  FO,  Fig.  5. 

845.  The    following   are  some   of    the 

properties  of  a  cone;   the  student  should  prove  the  correct- 
ness of  each : 

(a)  The  elements  of  a  right  circular  cone  are  equal. 

(b)  TJie  axis  of  a  right  circular  cone  is  equal  to  its  altitude. 

(c)  A  straight  line  drawn  from  the  vertex  of  a  cone  to  any 
point  in  the  perimeter  of  its  base  is  an  element. 

846.  Note.     The  theorems  and  exercises  on  the  cone  that  follow  will 
be  limited  to  cases  in  which  the  bases  of  the  cones  are  circles,  though  not 
necessarily  to  circular  cones.     When  the  term  cone  is  used,  therefore,  it 
must  be  understood  to  mean  a  cone  with  circular  base.    See  also  §  834. 


Ex.  1374.  What  is  the  locus  of  all  points  2  inches  from  the  lateral 
surface,  and  2  inches  from  the  base,  of  a  right  circular  cone  whose  alti- 
tude is  12  inches  and  the  radius  of  whose  base  is  5  inches  ? 


390 


SOLID   GEOMETRY 
PROPOSITION  III.     THEOREM 


847.  Every  section  of  a  cone  made  by  a  plane  passing 
through  its  vertex  is  a  triangle.          V 


B 


Given   cone  V—AB  with  base  AB  and  section  VCD  made  by 
a  plane  through  V. 
To  prove    VCD  a  A. 


ARGUMENT 

1.  From  V  draw  str.  lines  to  C  and  D. 

2.  Then  the  lines  so  drawn  are  elements ; 

i.e.  they  lie  in  the  conical  surface. 

3.  But  these  lines  lie  also  in  plane  VCD. 

.-.  they  are  the  intersections  of  plane 
VCD  with  the  conical  surface,  and 
coincide  with  VC  and  VD,  respectively. 

Also  CD  is  a  str.  line. 

.-.  VC,  VD,  and  CD  are  str.  lines  and  VCD 
is  a  A.  Q.E.D. 


4. 


REASONS 

1.  §  54,  15. 

2.  §  845,  c. 

3.  §  603,  a. 

4.  §614. 


5.  §  616. 

6.  §92. 


Ex.  1375.  What  kind  of  triangle  in  general  is  the  section  of  a  cone 
through  the  vertex,  if  the  cone  is  oblique  ?  if  the  cone  is  a  right  circular 
cone  ?  Can  any  section  of  an  oblique  cone  be  perpendicular  to  the  base 
of  the  cone?  of  a  right  circular  cone  ?  Explain. 

Ex.  1376.  Find  the  locus  of  all  straight  lines  making  a  given  angle 
with  a  given  straight  line,  at  a  given  point  in  the  line.  What  will  this 
locus  be  if  the  given  angle  is  90°  ? 

Ex.  1377.  Find  the  locus  of  all  straight  lines  making  a  given  angle 
with  a  given  plane  at  a  given  point.  What  will  the  locus  be  if  the  given 
angle  is  90°  ? 


BOOK   VIII 


391 


PROPOSITION  IV.     THEOREM 

848.   Every  section  of  a  cone  made  by  a  plane  parallel 
to  its  base  is  a  circle.    (See  §  846.) 

V 


Given  CD  a  section  of  cone  V-AB  made  by  a  plane  II  base  AB. 
To  prove    section  CD  a  O. 

OUTLINE  OF  PROOF 

1.  Let  R  and  S  be  any  two  points  on  the  boundary  of  section 
CD;  pass  planes  through  OF  and  points  R  and  S. 

2.  Prove  A  VOM  ~  A  VPR  and  A  VON  ~  A  VPS. 


3.   Then  «£=!2  and 


ON  _  VO       .       OM  _  ON 
PR       VP  PS  ~  VP  '  PR  ~  PS 

4.  But  OM=  ON-,  .-.  PS  =  PR;  i.e.  P  is  equidistant  from  any 
two  points  on  the  boundary  of  section  CD. 

5.  .-.  section  CD  is  a  O.  Q.E.D. 

849.   Cor.    Any  section  of  a  cone  parallel  to  its  base  is 
to  the  base  as  the  square  of  its  distance  from  the  vertex 
is  to  the  square  of  the  altitude  of  the  cone. 
OUTLINE  OF  PROOF 
section  CD     PR 


By  §  563, 

Prove 

Then 


base  AB 
PR 

OM 

section  CD 
base  AB 


VP 

vo 

VE 


VE 

VF 


For  applications  of  §§  848  and  849,  see  EXST  1380-1384. 


392  SOLID   GEOMETRY 

MENSURATION   OF   THE   CYLINDER   AND   CONE 

AREAS 

850.  Def.     A   plane  is  tangent  to  a   cylinder  if  it  contains 
an  element,  but  no  other  point,  of  the  cylinder. 

851.  Def.     A  prism  is  inscribed  in   a  cylinder  if  its  lateral 
edges  are  elements  of  the  cylinder,  and  the  bases  of  the  two 
figures  lie  in  the  same  plane. 

852.  Def.      A  prism  is  circumscribed  about  a  cylinder  if  its 
lateral  faces  are  all  tangent  to  the  cylinder,  and  the  bases  of 
the  two  figures  lie  in  the  same  plane. 


Ex.  1378.  How  many  planes  can  be  tangent  to  a  cylinder  ?  If  two 
of  these  planes  intersect,  the  line  of  intersection  is  parallel  to  an  element. 
How  are  the  bases  of  the  cylinders  in  §§  850-852  restricted  ?  (See  §  834.) 


853.  Before  proceeding  further  it  might  be  well  for  the 
student  to  review  the  more  important  steps  in  the  develop- 
ment of  the  area  of  a  circle.  In  that  development  it  was 
shown  that : 

(1)  The  area  of  a  regular  polygon  circumscribed   about  a 
circle  is  greater,  and  the  area  of  a  regular  polygon  inscribed  in 
a  circle  less,  than  the  area  of  the  regular  circumscribed  or  in- 
scribed polygon  of  twice  as  many  sides  (§  541). 

(2)  By  repeatedly  doubling  the  number  of  sides  of  regular 
circumscribed  and  inscribed  polygons  of  the  same  number  of 
sides,  and  making  the  polygons  always  regular,  their  areas 
approach  a  common  limit  (§  546). 

(3)  This  common  limit  is  defined  as  the  area  of  the  circle 
(§  558). 

(4)  Finally  follows  the  theorem  for  the  area  of  the  circle 
(§  559). 

It  will  be  observed  that  precisely  the  same  method  is  used 
throughout  the  mensuration  of  the  cylinder  and  the  cone. 

Compare  carefully  the  four  articles  just  cited  with  §§  854, 
855,  857,  and  858. 


BOOK  VIII  393 

PROPOSITION  V.     THEOREM 

854.  I.  The  lateral  area  of  a  regular  prism  circum- 
scribed about  a  right  circular  cylinder  is  greater  than 
the  lateral  area  of  the  regular  circumscribed  prism 
whose  base  has  twice  as  many  sides. 

II.  The  lateral  area  of  a  regular  prism  inscribed  in  a 
right  circular  cylinder  is  less  than  the  lateral  area  of  the 
regular  inscribed  prism  whose  base  has  twice  as  many 
sides. 


The  proof  is  left  as  an  exercise  for  the  student. 
HINT.     See  §  541.     Let  the  given  figures  represent  the  bases  of  the 
actual  figures. 

Ex.  1379.  A  regular  quadrangular  and  a  regular  octangular  prism 
are  inscribed  in  a  right  circular  cylinder  with  altitude  25  inches  and 
radius  of  base  10  inches.  Find  the  difference  between  their  lateral  areas. 

Ex.  1380.  The  line  joining  the  vertex  of  a  cone  to  the  center  of  the 
base,  passes  through  the  center  of  every  section  parallel  to  the  base. 

Ex.  1381.  Sections  of  a  cone  made  by  planes  parallel  to  the  base 
are  to  each  other  as  the  squares  of  their  distances  from  the  vertex. 

Ex.  1382.  The  base  of  a  cone  contains  144  square  inches  and  the 
altitude  is  10  inches.  Find  the  area  of  a  section  of  the  cone  3  inches 
from  the  vertex  ;  5  inches  from  the  vertex. 

Ex.  1383.  The  altitude  of  a  cone  is  12  inches.  How  far  from  the 
vertex  must  a  plane  be  passed  parallel  to  the  base  so  that  the  section 
shall  be  one  half  as  large  as  the  base  ?  one  third  ?  one  nth  ? 

Ex.  1384.  The  altitude  of  a  cone  is  20  inches  ;  the  area  of  the  section 
parallel  to  the  base  and  12  inches  from  the  vertex  is  90  square  inches. 
Find  the  area  of  the  base. 


394 


SOLID   GEOMETRY 


PROPOSITION  VI.     THEOREM 

855.  By  repeatedly  doubling  the  number  of  sides  of 
the  bases  of  regular  prisms  circumscribed  about,  and  in- 
scribed in,  a  right  circular  cylinder,  and  malting  the  bases 
always  regular  polygons,  their  lateral  areas  approach 
a  common  limit. 


Given  H  the  common  attitude,  R  and  r  the  apothems  of  the 
bases,  P  and  p  the  perimeters  of  the  bases,  and  S  and  s  the 
lateral  areas,  respectively,  of  regular  circumscribed  and  in- 
scribed prisms  whose  bases  have  the  same  number  of  sides. 
Let  the  given  figure  represent  the  base  of  the  actual  figure. 

To  prove  that  by  repeatedly  doubling  the  number  of  sides 
of  the  bases  of  the  prisms,  and  making  them  always  regular 
polygons,  S  and  s  approach  a  common  limit. 


1. 
2. 

3. 
4. 
5. 


ARGUMENT 

S  =  P  •  H  and  S  =  p  -  H. 

S_P 

s     p' 


p 

s 

s 
S-s 


r 
R 
r 
R  —  r 


REASONS 

1.  §  763. 

2.  §  54,  8  a. 

3.  §  538. 

4.  §  54,  1. 

5.  §  399. 


BOOK  VIII 


395 


6. 


8. 


9. 


11. 


—  s=  S 


ARGUMENT 

R  —  r 
B 


But  by  repeatedly  doubling  the  num- 
ber of  sides  of  the  bases  of  the 
prisms,  and  making  them  always 
regular  polygons,  R  —  r  can  be  made 
less  than  any  previously  assigned 
value,  however  small. 

.\  -        -    can  be  made   less  than  any 

previously  assigned  value,  however 
small. 
R 


S 


R 


can  be  made  less  than  any 


previously  assigned  value,  however 
small,  S  being  a  decreasing  variable. 

10.    .-.  S  —  s,  being  always  equal  to  S  —  — , 


can  be  made  less  than  any  previously 
assigned  value,  however  small. 
S  and  s  approach  a  common  limit. 

Q.E.D. 


REASONS 
6.    §  54,  7  a. 
7. '  §  543,  I. 


8.    §  586. 


9.   §  587. 


10.   §309. 


11.    §  594. 


856.  Note.     The  above  proof  is  limited  to  regular  prisms,  but  it  can 
be  shown  that  the  limit  of  the  lateral  area  of  any  inscribed  (or  circum- 
scribed) prism  is  the  same  by  whatever  method  the  number  of  the  sides 
of  its  base  is  successively  increased,  provided  that  each  side  approaches 
zero  as  a  limit.     (See  also  §  549.)     Compare  the  proof  of  §  855  with  that 
of  §  546,  I. 

857.  Def.      The  lateral  area  of  a  right  circular  cylinder  is  the 

common  limit  which  the  successive  lateral  areas  of  circum- 
scribed and  inscribed  regular  prisms  (having  bases  containing 
3,  4,  5,  etc.,  sides)  approach  as  the  number  of  sides  of  the 
bases  is  successively  increased  and  each  side  approaches  zero 
as  a  limit. 


396 


SOLID   GEOMETRY 


PROPOSITION  VII.     THEOREM 

858.  The  lateral  area  of  a  right  circular  cylinder  is 
equal  to  the  product  of  the  circumference  of  its  base  and 
its  altitude. 


Given  a  rt.  circular  cylinder  with  its  lateral  area  denoted  by 
,  the  circumference  of  its  base  by  (7,  and  its  altitude  by  //. 
To  prove    S  =  C  •  H. 


ARGUMENT 

1.  Circumscribe    about    the    rt.     circular 

cylinder  a  regular  prism.  Denote 
its  lateral  area  by  S',  the  perimeter 
of  its  base  by  P,  and  its  altitude 
by  H. 

2.  Then  S'  =  P  -  H. 

3.  As  the  number  of  sides  of  the  base  of 

the  regular  circumscribed  prism  is 
repeatedly  doubled,  P  approaches  O 
as  a  limit. 

4.  .-.  P  -  H  approaches  C  •  H  as  a  limit. 

5.  Also  S'  approaches  S  as  a  limit. 
G.    But  S1  is  always  equal  to  P  •  H. 

7.    .-.  8=  C  -  H.  Q.E.D. 


REASONS 


1.    §  852. 


2.  §  763. 

3.  §  550. 


4.  §  590. 

5.  §857. 

6.  Arg.  2. 

7.  §  355. 


859.  Cor.  //  S  denotes  the  lateral  area,  T  the  total  area, 
H  the  altitude,  and  E  the  radius  of  the  base,  of  a  right 
circular  cylinder, 


BOOK  VIII 


397 


T= 


=  2 


860.    Note.     Since  the  lateral  area  of  an  oblique  prism  is  equal  to 
the  product  of  the  perimeter  of  a  right  section  and  a  lateral  edge  (§  762), 
the  student  would  naturally  infer  that  the  lateral 
area  of  an  oblique   cylinder  with  circular  bases  is 
equal  to  the  product  of  the  perimeter  of  a  right  sec- 
tion and  an  element.     This  statement  is  true.     But 
the  right  section  of  an  oblique  cylinder  with  circular 
base  is  not  a  circle.     And  since  the  only  curve  dealt 
with  in   elementary    geometry  is.  the    circle,   this 
theorem  and  its  applications  have  been  omitted  here. 


Ex.  1385.  Find  the  lateral  area  and  total  area  of  a  right  circular 
cylinder  whose  altitude  is  20  centimeters  and  the  diameter  of  whose  base 
is  10  centimeters. 

Ex.  1386.  How  many  square  inches  of  tin  will  be  required  to  make 
an  open  cylindrical  pail  8  inches  in  diameter  and  10  inches  deep,  making 
no  allowance  for  waste  ? 

Ex.  1387.  In  a  right  circular  cylinder,  find  the  ratio  of  the  lateral 
area  to  the  sum  of  the  two  bases.  What  is  this  ratio  if  the  altitude  and 
the  radius  of  base  are  equal  ? 

Ex.  1388.  Find  the  altitude  of  a  right  circular  cylinder  if  its  lateral 
area  is  8  and  the  radius  of  its  base  B. 

Ex.  1389.  Find  the  radius  of  the  base  of  a  right  circular  cylinder  if 
its  total  area  is  T  and  its  altitude  is  H. 


861.  Def.     Because  it  may  be  generated  by  a  rectangle  re- 
volving  about   one   of   its   sides  as  an  axis,  a  right   circular 
cylinder  is  sometimes  called  a  cylinder  of  rev- 
olution. 

862.  Questions.     What  part  of  the  cylinder  will 
side  CD,  opposite  the  axis  AB,  generate?     What  will 
AD  and  BC  generate  ?    What  will  the  plane  AC  gen- 
erate ?    What  might  CD  in  any  one  of  its  positions  be 
called  ? 

863.  Def.      Similar    cylinders    of   revolution 

are  cylinders  generated  by  similar  rectangles  revolving  about 
homologous  sides  as  axes. 


398 


SOLID   GEOMETRY 


PROPOSITION  VIII.     THEOREM 

864.  The  lateral  areas,  and  the  total  areas,  of  two 
similar  cylinders  of  revolution  are  to  each,  other  as  the 
squares  of  their  altitudes,  and  as  the  squares  of  the  radii 
of  their  bases. 


Given  two  similar  cylinders  of  revolution  with  their  lateral 
areas  denoted  by  S  and  s',  their  total  areas  by  T  and  T1,  their 
altitudes  by  IT  and  H',  and  the  radii  of  their  bases  by  .Rand  Rr, 
respectively. 

To  prove:    (.)        -- 


H' 


R'2 


ARGUMENT 
=  27rRH  and  Sf  =  2  -rrR'H'. 

2<rrRH  _     RH  _  R      JJ 
~  R1  '  Hf 


1. 


2.  .-.  —  = 

Sf      2 

3.  But  rectangle  RH  ~  rectangle  RfH/. 

H       R 

4-  •'•?-*• 

K       .    _^._^       H  _H2 
'  'Sr~  H'  '  H'~  H'*' 

S'  ~~  R'   '  R'  ~~  R'2 
7.  Again,  r=  2  7r.R(#-f  /?) 

and  T'  =  2  Tr^Yfl"'  4- 


REASONS 

1.  §  859. 

2.  §  54,  8  a. 

3.  §  863. 

4.  §  419. 

5.  §  309. 

6.  §  309. 

7.  §  859. 


BOOK  Ylll 


399 


ARGUMENT 


H+R 


T'       R'(H'  +  R')       R'      H'  +  tf 

H+R         H       R 


9.    But,  from  Arg.  4, 


Hf 


10      •  —  =  —  •  —  =  — 
T'      //'     /?'     //'^ 

r      i?     ^      R2 

11.  Also  —  =  — -  •  —-  =  —-.« 

T'      R>     #      R'2 


Q.E.D. 


REASONS 

8.  §  54,  8  a. 

9.  §  401. 

10.  §  309. 

11.  §  309. 


Ex.  1390.  The  altitudes  of  two  similar  cylinders  of  revolution  are  5 
inches  and  7  inches,  respectively,  and  the  total  area  of  the  first  is  676 
square  inches.  Find  the  total  area  of  the  second. 

Ex.  1391.  The  lateral  areas  of  two  similar  cylinders  of  revolution  are 
320  square  inches  and  500  square  inches,  and  the  radius  of  the  base  of 
the  larger  is  10  inches.  Find  the  radius  of  the  base  of  the  smaller. 

Ex.  1392.  Two  adjacent  sides  of  a  rectangle  are  a  and  b;  find 
the  lateral  area  of  the  cylinder  generated  by  revolving  the  rectangle  :  (1) 
about  a  as  an  axis  ;  (2)  about  b  as  an  axis.  Put  the  results  in  the  form 
of  a  general  statement.  Have  you  proved  this  general  statement  ? 


865.  Def.     The  slant  height  of  a  right  circular  cone  is   a 
straight  line  joining  its  vertex  to  any  point  in  the  circumference 
of  its  base.     Thus  any  element  of  such  a  cone  is  its  slant  height. 

866.  Def.     A  plane  is  tangent  to  a  cone  if  it  contains  an 
element,  but  no  other  point,  of  the  cone. 

867.  Def.     A  pyramid  is  inscribed  in  a  cone  if  its  base  is 
inscribed  in  the  base  of  the  cone  and  its  vertex  coincides  with 
the  vertex  of  the  cone. 

868.  Def.     A  pyramid  is  circumscribed  about  a  cone  if  its 
base   is   circumscribed   about   the  base   of   the   cone  and  its 
vertex  coincides  with  the  vertex  of  the  cone. 

869.  The  student  may  state  and  prove  the  theorems  on  the 
right  circular  cone  corresponding  to  those  mentioned  in  §  854. 

Ex.  1393.  How  many  planes  can  be  tangent  to  a  cone  ?  Through 
what  point  must  each  of  these  planes  pass  ?  Prove.  How  are  the  bases 
of  the  cones  in  §§  860-868  restricted  ?  (See  §  846.) 


400- 


SOLID  GEOMETRY 


PROPOSITION  IX.     THEOREM 

870.  By  repeatedly  doubling  the  number  of  sides  of  the 
bases  of  regular  pyramids  circumscribed  about,  and  in- 
scribed in,  a  right  circular  cone,  and  making  the  bases 
always  regular  polygons,  their  lateral  areas  approach 
a  common  limit. 


Given  H  the  common  altitude,  L  and  I  the  slant  heights,  R 
and  r  the  apothems  of  the  bases.  P  and  p  the  perimeters  of  the 
bases,  and  S  and  s  the  lateral  areas,  respectively,  of  regular 
circumscribed  and  inscribed  pyramids  whose  bases  have  the 
same  number  of  sides.  Let  the  given  figure  represent  the 
base  of  the  actual  figure. 

To  prove  that  by  repeatedly  doubling  the  number  of  sides 
of  the  bases  of  the  pyramids,  and  making  them  always  regular 
polygons,  S  and  s  approach  a  common  limit. 


ARGUMENT 

1.    S  =  1  PL  and  s  =  1  pi. 
p 


-       ~     - 
s      pi      p      I 


2 


3. 


i  O  / 1  J  J  1\  LJ 

'  s      r      I       rl' 
—  s      RL  —  rl 


5.    .-. 


RL 


REASONS 

1.  §  766. 

2.  §  54,  8  a. 

3.  §  538. 

4.  §  309. 

5.  §  399. 


BOOK  VIII 


401 


ARGUMENT 


9. 


10. 
11. 
12. 
13. 


14. 


15. 
16. 
17. 


,_.  = 


RL 

Now  L  —  l<CB. 

But  by  repeatedly  doubling  the  num- 
ber of  sides  of  the  bases  of  the 
pyramids,  and  making  them  always 
regular  polygons,  CB  can  be  made 
less  than  any  previously  assigned 
value,  however  small. 

.*.  L  —  l)  being  always  less  than  CB,  can 
be  made  less  than  any  previously 
assigned  value,  however  small. 

.-.  the  limit  of  I  =  L. 

Also  the  limit  of  r  =  R. 

.•.  the  limit  of  rl  =  RL. 

.-.  RL  —  rl  can  be  made  less  than  any 
previously  assigned  value,  however 
small. 
RL  -  rl 


RL 


can  be  made  less  than  any 


previously  assigned  value,  however 

small. 

Similar  to  Arg.  9,  §  855. 
Similar  to  Arg.  10,  §  855. 
.-.  S  and  s  approach  a  common  limit. 

Q.E.D. 


REASONS 

6.  54,  7  a. 

7.  §  168. 

8.  Arg.  3,  §543, 

I. 


9.    §  54,  10. 


10.  §  349. 

11.  §543,1. 

12.  §  592. 

13.  §349. 


14.    §  586. 


15.  §  587. 

16.  §  309. 

17.  §  594. 


871.    Note.     The  proof  of  §  870  is  limited  in  the  same  manner  as  the 
proof  of  §  855.     Read  §  856. 


872.    Def.     The  lateral  area  of   a  right  circular  cone  is  the 

common  limit  which  the  successive  lateral  areas  of  circum- 
scribed and  inscribed  regular  pyramids  approach  as  the  num- 
ber of  sides  of. the  bases  is  successively  increased  and  each 
side  approaches  zero  as  a  linm. 


402  SOLID   GEOMETRY 

PROPOSITION  X.     THEOREM 

873.  The  lateral  area  of  a  right  circular  cone  is  equal 
to  one  half  the  product  of  the  circumference  of  its  base 
and  its  slant  height. 


Given  a  rt.  circular  cone  with  its  lateral  area  denoted  by  S, 
the  circumference  of  its  base  by  (7,  and  its  slant  height  by  L. 
To  prove  S  =  \  C  -  L. 
The  proof  is  left  as  an  exercise  for  the  student. 

874.  Question.      What  changes  are  necessary  in  the  proof  of  Prop. 
VII  to  make  it  the  proof  of  Prop.  X  ? 

875.  Cor.     If  s  denotes  the  lateral  area,  T  the  total  area, 
L  the  slant  height,  and  R  the  radius  of  the  base,  of  a  right 
circular  cone,       s  =  irRL ;  • 

T  =  TrRL  +  irR2  =  irR(L  +  #). 


Ex.  1394.  The  altitude  of  a  right  circular  cone  is  12  inches  and  the 
radius  of  the  base  8  inches.  Find  the  lateral  area  and  the  total  area  of 
the  cone. 

Ex.  1395.  How  many  yards  of  canvas  30  inches  wide  will  be  required 
to  make  a  conical  tent  16  feet  high  and  20  feet  in  diameter,  if  10  %  of  the 
goods  is  allowed  for  cutting  and  fitting  ? 

Ex.  1396.  The  lateral  area  of  a  right  circular  cone  is  ^|^V89  square 
inches,  and  tke  radius  of  the  base  is  10  inches.  Find  the  altitude. 


876.  Def.    Because  it  may  be  generated  by  a  right  triangle 
revolving  about  one  of  its  sides  as  an  axis,  a  right  circular  cone 
is  sometimes  called  a  cone  of  revolution. 

877.  Def.     Similar  cones  of  revolution  are  cones  generated  by 
similar  right  triangles  revolving  about  homologous  sides  as  axes. 


BOOK  VIII  403 


PROPOSITION  XI.     THEOREM 

878.  The  lateral  areas,  and  the  total  areas,  of  two  simi- 
lar cones  of  revolution  are  to  each  other  as  the  squares  of 
their  altitudes,  as  the  squares  of  their  slant  heights, 
and  as  the  squares  of  the  radii  of  their  bases. 


Given  two  similar  cones  of  revolution  with  their  lateral 
areas  denoted  by  S  and  S',  their  total  areas  by  T  and  T\  their 
altitudes  by  H  and  H1,  their  slant  heights  by  L  and  L1,  and 
the  radii  of  their  bases  by  R  and  R1,  respectively. 

,  N     S        H2         L2        R2 
Toprove:  (a)    -  =  _  =  -  =  -, 

(6)     r=^=^=^ 

V    )      T,          ff,2          L,2          R,2 

-The  proof  is  left  as  an  exercise  for  the  student. 
HINT.    Apply  the  method  of  proof  used  in  Prop.  VIII. 


879.  Def.     A  frustum  of  a  cone  is  the  portion  of  the  cone 
included  between  the  base  and  a  section  of  the  cone  made  by 
a  plane  parallel  to  the  base. 

880.  Questions.     What  are  the  upper  and  lower  bases  of  a  frustum 
of  a  cone  ?  the  altitude  ?     What  kind  of  a  figure  is  the  upper  base  of  a 
frustum  of  a  right  circular  cone  (§  848)  ? 

881.  Def.     The  slant  height  of  a  frustum  of  a  right  circular 
cone  is  the  length  of  that  portion  of  an  element  of  the  cone 
included  between  the  bases  of  the  frustum. 


Ex.  1397.     Every  section  of  a  frustum  of  a  cone,  made  by  a  plane 
passing  through  an  element,  is  a  trapezoid. 


404 


SOLID   GEOMETRY 


PROPOSITION  XII.     THEOREM 

882.  The  lateral  area  of  a  frustum  of  a  right  circular 
cone  is  equal  to  one  half  the  product  of  the  sum  of  the 
circumferences  of  its  bases  and  its  slant  height. 


M 


Given  frustum  AM,  of  right  circular  cone  0-AF,  with  its 
lateral  area  denoted  by  S,  the  circumferences  of  its  bases  by 
C  and  c,  the  radii  of  its  bases  by  R  and  r}  and  its  slant  height 
by  L. 

To  prove  S  =  l((7-j-c)L. 

ARGUMENT 

1.  S  =  lateral   area  of   cone   0-AF  minus 

lateral  area  of  cone  O-DM. 

2.  Let  L'  denote  the  slant  height  of  cone 

O-DM.    Then  S  =  |  c(L  +  L')  —  \  cL' 


3. 


It  now  remains  to  find  the  value  of  L'. 
C_R 
c      r 
4.   But  A  OKD  ~  A  OQA. 

5      •    *  —  OA  —  L  +  L' 
r~OD~ 


L1 


L'  = 


CL 

C-c 


8. 


cL 

1 

2— c 


Q.E.D. 


REASONS 

1.  §  54,  11. 

2.  §  873. 


3.  §556. 

4.  §  422. 

5.  §  424,  2. 

6.  §54,  1. 

7.  Solving  for  //. 

8.  §  309. 


BOOK  VIII  405 

883.  Cor.  I.  If  s  denotes  the  lateral  area,  T  the  total 
area,  L  the  slant  height,  and  R  and  r  the  radii  of  the 
bases,  of  a  frustum  of  a  right  circular  cone, 


884.  Cor.  n.  The  lateral  area  of  a  frustum  of  a  right 
circular  cone  is  equal  to  the  product  of  its  slant  height 
and  the  circumference  of  a  section  midway  between  its 
bases. 

Ex.  1398.  If  8  denotes  the  lateral  area,  L  the  slant  height,  and  O 
the  circumference  of  a  section  midway  between  the  bases,  of  a  frustum  of 
a  right  circular  cone,  then  S  =  CL. 

Ex.  1399.  In  the  formulas  of  §  883  :  (a)  make  r=  0  and  compare 
results  with  formulas  of  §  875  ;  (6)  make  b  =  B  and  compare  results 
with  formulas  of  §  859. 

Ex.  1400.  The  altitude  of  a  frustum  of  a  right  circular  cone  is 
16  inches,  and  the  diameters  of  its  bases  are  20  inches  and  30  inches, 
respectively.  Find  its  lateral  area  and  also  its  total  area. 

Ex.  1401.     In  the  figure,   AB  and   CD  are  arcs  of  circles;   OA  = 
2  inches,  OD  =  5  inches,  and  ZDOC=  120°.     Cut  figure  ABCD  out  of 
paper  and  form  it  into  a  frustum  of  a  cone.     Find 
its  lateral  area  and  also  its  total  area. 

Ex.  1402.  A  frustum  of  a  right  circular  cone 
whose  altitude  is  4  inches  and  radii  of  bases  4 
inches  and"  7  inches,  respectively,  is  made  as  in- 
dicated in  Ex.  1401.  Find  the  radius  of  the  circle 
from  which  it  must  be  cut. 

Ex.  1403.  The  sum  of  the  total  areas  of  two  similar  cylinders  of  rev- 
olution is  216  square  inches,  and  one  altitude  is  f  of  the  other.  Find  the 
total  area  of  each  cylinder. 

Ex.  1404.  A  regular  triangular  and  a  regular  hexangular  pyramid 
are  inscribed  in  a  right  circular  cone  with  altitude  20  inches  and  with 
radius  of  base  4  inches.  Find  the  difference  between  their  lateral  areas. 

Ex.  1405.  Cut  out  of  paper  a  semicircle  whose  radius  is  4  inches, 
and  find  its  area.  Form  a  cone  with  this  semicircle  and  find  its  lateral 
area  by  §  875.  Do  the  two  results  agree  ? 

Ex.  1406.  The  slant  height,  and  the  diameter  of  the  base,  of  a  right 
circular  cone  are  each  equal  to  L.  Find  the  total  area. 


406 


SOLID   GEOMETRY 


VOLUMES 
PROPOSITION  XIII.     THEOREM 

885.  I.  The  volume  of  a  prism  whose  base  is  a  regular 
polygon  and  which  is  circumscribed  about  a  cylinder  is 
greater  than  the  volume  of  the  circumscribed  prism, 
whose  base  is  a  regular  polygon  with  twice  as  many 
sides. 

II.  The  volume  of  a  prism  whose  base  is  a  regular  poly- 
gon and  which  is  inscribed  in  a  cylinder  is  less  than  the 
volume  of  the  inscribed  prism  whose  base  is  a  regular 
polygon  with  twice  as  ?nany  sides. 

The  figures  and  proofs  are  left  as  exercises  for  the  student. 


PROPOSITION  XIV.     THEOREM 

886.  By  repeatedly  doubling  the  number  of  sides  of  the 
bases  of  prisms  circumscribed  about,  and  inscribed  in, 
a  cylinder,  and  making  the  bases  always  regular  poly- 
gons, their  volumes  approach  a  common  limit. 


Given  H  the  common  altitude,  B  and  b  the  areas  of  the 
bases,  and  v  and  v  the  volumes,  respectively,  of  circumscribed 
and  inscribed  prisms  whose  bases  are  regular  and  have  the 
same  number  of  sides.  Let  the  given  figure  represent  the  base 


of  the  actual  figure. 


BOOK   VIII 


407 


To  prove  that  by  repeatedly  doubling  the  number  of  sides  of 
the  bases  of  the  prisms,  and  making  them  always  regular  poly- 
gons, V  and  v  approach  a  common  limit. 


ARGUMENT 
1.    F  =  B  •  H  and  v  =  b  •  H. 

2      •  V  =B  '  H  =  B 
b' 


v       b  •  H 

V—v      B—b 


B 


4. 


V  —  v  = 


B 


5.  Similar  to  Arg.  7,  §  855. 

6.  Similar  to  Arg.  8,  §  855. 

7.  Similar  to  Arg.  9,  §  855. 

8.  Similar  to  Arg.  10,  §  855. 

9.  /.  rand  v  approach  a  common  limit. 

Q.E.D. 


REASONS 

1.  §  799. 

2.  §  54,  8  a. 

3.  §  399. 

4.  §  54,  7  a. 

5.  §  546,  II. 

6.  §  586. 

7.  §  587. 

8.  §  309. 

9.  §  594. 


887.    Note.     The  proof  of  §  886  is  limited  in  the  same  manner  as  the 
proof  of  §  855.     Read  §  856. 


Ex.  1407.  The  total  area  of  a  right  circular  cone  whose  altitude  is 
10  inches  is  280  square  inches.  Find  the  total  area  of  the  cone  cut  off  by 
a  plane  parallel  to  the  base  and  6  inches  from  the  base. 

Ex.  1408.  The  altitude  of  a  right  circular  cone  is  12  inches.  What 
part  of  the  lateral  surface  is  cut  off  by  a  plane  parallel  to  the  base  and 
6  inches  from  the  vertex  ? 

Ex.  1409.  The  altitude  of  a  right  circular  cone  is  //.  How  far  from 
the  vertex  must  a  plane  be  passed  parallel  to  the  base  so  that  the  lateral 
area  and  the  total  area  of  the  cone  cut  off  shall  be  one  half  that  of  the 
original  cone  ?  one  third  ?  one  nth  ? 


888.  Def.  The  volume  of  a  cylinder  is  the  common  limit 
which  the  successive  volumes  of  circumscribed  and  inscribed 
prisms  approach  as  the  number  of  sides  of  the  bases  is  suc- 
cessively increased,  and  each  side  approaches  zero  as  a  limit. 


408 


SOLID   GEOMETRY 


PROPOSITION   XV.     THEOREM 

889.    The  volume  of  a  cylinder  is  equal  to  the  product 
of  its  base  and  its  altitude. 


Given  a  cylinder,  with  its  volume  denoted  by  F,  its  base  by 
,  and  its  altitude  by  H. 
To  prove  F  =  B  •  H. 


ARGUMENT 

1.  Circumscribe   about    the   cylinder   a 

prism  whose  base  is  a  regular  poly- 
gon. Denote  its  volume  by  V1  and 
the  area  of  its  base  by  B'. 

2.  Then  Ff  =  B1  •  H. 

3.  As  the  number  of  sides  of  the  base 

of  the  circumscribed  prism  is  re- 
peatedly doubled,  B1  approaches  B 
as  a  limit. 

4.  .•.  B1  •  H  approaches  B  •  H  as  a  limit. 

5.  Also  F'  approaches  V  as  a  limit. 

6.  But  F'  is  always  equal  to  B'  •  H. 

7.  .-.  V=B  •  H.  Q.E.D. 


REASONS 
1.    §  852. 


2.  §  799. 

3.  §558. 


4.  §  590. 

5.  §  888. 

6.  Arg.  2. 

7.  §  355. 


890.   Cor.     //  F  denotes  the  volume,  H  the  altitude,  and 
R  the  radius  of  the  base,  of  a  cylinder, 


BOOK  VIII 


409 


PROPOSITIOX  XVI.     THEOREM 

891.  The  volumes  of  two  similar  cylinders  of  revolution 
are  to  each  other  as  the  cubes  of  their  altitudes,  and  as 
the  cubes  of  the  radii  of  their  bases. 


Given  two  similar  cylinders  of  revolution  with  their  volumes 
denoted  by  F  and  F;,  their  altitudes  by  H  and  H1,  and  the  radii 
of  their  bases  by  R  and  R',  respectively. 


F       Hs      R3 
To  prove  — -  =  — -  =  — . 

yl         HI3         RI3 


ARGUMENT 

1.     F  =  7rR2H  and  V1  =  7rR'2H'. 
2       -    F  =  7rR2H  _  R2H  =  R2     H_ 
'    V'       7rR'2H'       RI2H'~  R12'  H1' 
3.    But  rectangle  RH  ~  rectangle  R'H1. 

4      •  IL=R- 
'  H'       R'' 


5. 


R^ 

>'2 


R'2    R'       R'3 
6.    But,  from  Arg.  4,  f-  =  — . 


7.    ...JU* 

Ff       ^ 


_ 

'3' 


Q.  E.  D. 


REASONS 

1.  §890. 

2.  §  54,  8  a. 

3.  §  863. 

4.  §  419. 

5.  §  309. 

6.  §  54,  13. 

7.  §  309. 


Ex.  1410.  The  volumes  of  two  similar  cylinders  of  revolution  are 
135  cubic  inches  and  1715  cubic  inches,  respectively,  and  the  altitude  of 
the  first  is  3  inches.  Find  the  altitude  of  the  second. 


410  SOLID   GEOMETRY 

Ex.  1411.  A  cylinder  of  revolution  has  an  altitude  of  12  inches  and 
a  base  with  a  radius  of  5  inches.  Find  the  total  area  of  a  similar  cylinder 
whose  volume  is  8  times  that  of  the  given  cylinder. 

Ex.  1412.  The  dimensions  of  a  rectangle  are  6  inches  and  8  inches, 
respectively.  Find  the  volume  of  the  solid  generated  by  revolving  the 
rectangle :  (a)  about  its  longer  side  as  an  axis  ;  (6)  about  its  shorter  side. 
Compare  the  ratio  of  these  volumes  with  the  ratio  of  the  sides  of  the 
rectangle. 

Ex.  1413.  Cylinders  having  equal  bases  and  equal  altitudes  are 
equivalent. 

Ex.  1414.  Any  two  cylinders  are  to  each  other  as  the  products  of 
their  bases  and  their  altitudes. 

Ex.  1415.  (a)  Two  cylinders  having  equal  bases  are  to  each  other 
as  their  altitudes,  and  (&)  having  equal  altitudes  are  to  each  other  as 
their  bases. 

Ex.  1416.  The  volume  of  a  right  circular  cylinder  is  equal  to  the 
product  of  its  lateral  area  and  one  half  the  radius  of  its  base. 

Ex.  1417.  Cut  out  a  rectangular  piece  of  paper  0x9  inches.  Roll 
this  into  a  right  circular  cylinder  and  find  its  volume  (two  answers). 

Ex.  1418.  A  cistern  in  the  form  of  a  right  circular  cylinder  is  to  be 
20  feet  deep  and  8  feet  in  diameter.  How  much  will  it  cost  to  dig  it  at 
5  cents  a  cubic  foot  ? 

Ex.  1419.  Find  the  altitude  of  a  right  circular  cylinder  if  its  volume 
is  Fand  the  radius  of  its  base  R. 

Ex.  1420.  In  a  certain  right  circular  cylinder  the  lateral  area  and 
the  volume  have  the  same  numerical  value,  (a)  Find  the  radius  of  the 
base.  (&)  Find  the  volume  if  the  altitude  is  equal  to  the  diameter  of  the 
base. 

Ex.  1421.  A  cylinder  is  inscribed  in  a  cube  whose  edge  is  10  inches. 
Find  :  (a)  the  volume  of  each  ;  (ft)  the  ratio  of  the  cylinder  to  the  cube. 

Ex.  1422.  A  cylindrical  tin  tomato  can  is  4^  inches  high,  and  the 
diameter  of  its  base  is  4  inches.  Does  it  hold  more  or  less  than  a  liquid 
quart,  i.e.  ^f-1  cubic  inches  ? 

892.   The  student  may : 

(a)  State  and  prove  the  theorems  on  the  cone  corresponding 
to  those  given  in  §§  885  and  886. 

(b)  State,  by  aid  of  §  888,  the  definition  of  the  volume  of  a 
cone. 


BOOK  VIII  411 


PROPOSITION  XVII.     THEOREM 

893.   The  volume  of  a  cone  is  equal  to  one  third  the 
product  of  its  base  and  its  altitude. 


Given  a  cone  with  its  volume  denoted  by  F,  its  base  by  B, 
and  its  altitude  by  H. 
To  prove    V  =  J  B  -  H. 
The  proof  is  left  as  an  exercise  for  the  student. 

894.  Question.     What  changes  must  be  made  in  the  proof  of  Prop. 
XV  to  make  it  the  proof  of  Prop.  XVII  ? 

895.  Cor.     If  V  denotes  the  volume,  H  the  altitude,  and 
R  the  radius  of  tlie  base  of  a  cone, 


Ex.  1423.  Any  two  cones  are  to  each  other  as  the  products  of  their 
bases  and  altitudes. 

Ex.  1424.  The  slant  height  of  a  right  circular  cone  is  18  inches  and 
makes  with  the  base  an  angle  of  60° ;  the  radius  of  the  base  is  8  inches. 
Find  the  volume  of  the  cone. 

Ex.  1425.  The  base  of  a  cone  has  a  radius  of  12  inches  ;  an  element 
of  the  cone  is  24  inches  long  and  makes  with  the  base  an  angle  of  30°. 
Find  the  volume  of  the  cone. 

Ex.  1426.  The  hypotenuse  of  a  right  triangle  is  17  inches  and  one 
side  is  15  inches.  Find  the  volume  of  the  solid  generated  by  revolving  the 
triangle  about  its  shortest  side  as  an  axis. 

Ex.  1427.  A  cone  and  a  cylinder  have  equal  bases  and  equal  alti- 
tudes. Find  the  ratio  of  their  volumes. 


896.  Historical  Note.  To  Eudoxus  is  credited  the  proof  of  the 
proposition  that  "every  cone  is  the  third  part  of  a  cylinder  on  the  same 
base  and  with  the  same  altitude."  Proofs  of  this  proposition  were  also 
given  later  by  Archimedes  and  Brahmagupta.  (Compare  with  §  809. ) 


412  SOLID  GEOMETRY 

PROPOSITION  XVIII.     THEOREM 

897.  The  volumes  of  two  similar  cones  of  revolution 
are  to  each  other  as  the  cubes  of  their  altitudes,  as  the 
cubes  of  their  slant  heights,  and  as  the  cubes  of  the  ra- 
dii of  their  bases. 


Given  two  similar  cones  of  revolution  with  their  volumes 
denoted  by  V  and  V',  their  altitudes  by  H  and  H\  their  slant 
heights  by  L  and  L',  and  the  radii  of  their  bases  by  R  and  £', 
respectively. 

To  prove  Z*=4      * 

V1      H13      L'3      R'3 

The  proof  is  left  as  an  exercise  for  the  student. 
HINT.  Apply  the  method  of  proof  used  in  Prop.  XVI. 


Ex.  1428.  If  the  altitude  of  a  cone  of  revolution  is  three  fourths  that 
of  a  similar  cone,  what  other  fact  follows  by  definition  ?  Compare  the 
circumferences  of  the  two  bases ;  their  areas.  Compare  the  total  areas 
of  the  two  cones  ;  their  volumes. 

Ex.  1429.  If  the  lateral  area  of  a  right  circular  cone  is  1T9¥  times  that 
of  a  similar  cone,  what  is  the  ratio  of  their  volumes  ?  of  their  altitudes  ? 

Ex.  1430.  Through  a  given  cone  X  two  planes  are  passed  parallel  to 
the  base  ;  let  Y  denote  the  cone  cut  off  by  the  upper  plane,  and  Z  the 
entire  cone  cut  off  by  the  lower  plane.  Prove  that  Y  and  Z  are  to  each 
other  as  the  cubes  of  the  distances  of  the  planes  from  the  vertex  of  the 
given  cone  X. 

HINT.    See  Ex.  1381. 

Ex.  1431.   Show  that  §  897  is  a  special  case^f  Ex.  1430. 

Ex.  1432.  The  lateral  area  of  a  cone  of  revolution  is  144  square 
inches  and  the  total  area  240  square  inches.  Find  the  volume. 


BOOK  VIII  413 

PROPOSITION  XIX.     THEOREM 

898.  The  volume  of  a  frustum  of  a  cone  is  equal  to  one 
third  the  product  of  its  altitude  and  the  sum  of  its 
lower  base,  its  upper  base,  and  the  mean  proportional 

between  its  two  bases. 

0 


Given  frustum  AM,  of  cone  0-AF,  with  its  volume  denoted 
by  F,  its  lower  base  by  By  its  upper  base  by  6,  and  its 
altitude  by  H. 

To  prove      F=  -J- H(B  +  b  +  V^6). 

The  proof  is  left  as  an  exercise  for  the  student. 

HINT.     In  the  proof  of  §  815,  change  "pyramid"  to  "cone." 


Cor.     If  v  denotes  the  volume,  H  the  altitude,  and 
and  r  the  radii  of  the  bases  of  a  frustum  of  a  cone, 


Ex.  1433.  Make  a  frustum  of  a  right  circular  cone  as  indicated  in 
Ex.  1401,  and  of  the  same  dimensions.  Find  its  volume. 

Ex.  1434.  A  tin  pail  is  in  the  form  of  a  frustum  of  a  cone ;  the 
diameter  of  its  upper  base  is  12  inches,  of  its  lower  base  10  inches. 
How  high  must  the  pail  be  to  hold  2j  gallons  of  water  ?  (See  Ex.  1422. ) 

Ex.  1435.  A  cone  6  feet  high  is  cut  by  a  plane  parallel  to  the  base 
and  4  feet  from  the  vertex  ;  the  volume  of  the  frustum  formed  is  456 
cubic  inches.  Find  the  volume  of  the  entire  cone. 

Ex.  1436.  Find  the  ratio  of  the  base  to  the  lateral  area  of  a  right 
circular  cone  whose  altitude  is  equal  to  the  diameter  of  its  base. 


414  SOLID   GEOMETRY 


MISCELLANEOUS   EXERCISES 

Ex.  1437.  The  base  of  a  cylinder  is  inscribed  in  a  face  of  a  cube 
whose  edge  is  10  inches.  Find  the  altitude  of  the  cylinder  if  its  volume 
is  equal  to  the  volume  of  the  cube. 

Ex.  1438.  A  block  of  marble  in  the  form  of  a  regular  prism  is  10  feet 
long  and  2  feet  6  inches  square  at  the  base.  Find  the  volume  of  the 
largest  cylindrical  pillar  that  can  be  cut  from  it. 

Ex.  1439.  The  Winchester  bushel,  formerly  used  in  England,  was 
the  volume  of  a  right  circular  cylinder  18|  inches  in  internal  diameter 
and  8  inches  in  depth.  Is  this  the  same  volume  as  the  bushel  used  in  the 
United  States  (2150.42  cubic  inches)  ? 

Ex.  1440.  To  determine  the  volume  of  an  irregular  body,  it  was 
immersed  in  a  vessel  containing  water.  The  vessel  was  in  the  form  of  a 
right  circular  cylinder  the  radius  of  whose  base  was  8  inches.  On  placing 
the  body  in  the  cylinder,  the  surface  of  the  water  was  raised  10|  inches. 
Find  the  volume  of  the  irregular  solid. 

Ex.  1441.  In  draining  a  certain  pond  a  4-inch  tiling  (i.e.  a  tiling 
whose  inside  diameter  was  4  inches)  was  used.  In  draining  another 
pond,  supposed  to  contain  half  as  much  water,  a  2-inch  tiling  was  laid. 
It  could  not  drain  the  pond.  What  was  the  error  made  ? 

Ex.  1442.  A  grain  elevator  in  the  form  of  a  frustum  of  a  right  cir- 
cular cone  is  25  feet  high  ;  the  radii  of  its  bases  are  10  feet  and  5  feet, 
respectively  ;  how  many  bushels  of  wheat  will  it  hold,  counting  1|-  cubic 
feet  to  a  bushel  ? 

Ex.  1443.  The  altitude  of  a  cone  with  circular  base  is  16  inches. 
At  what  distance  from  the  vertex  must  a  plane  be  passed  parallel  to  the 
base  to  cut  the  cone  into  two  equivalent  parts  ? 

Ex.  1444.  Two  sides  of  a  triangle  including  an  angle  of  120°  are  10 
and  20,  respectively.  Find  the  volume  of  the  solid  generated  by  revolv- 
ing the  triangle  about  side  10  as  an  axis. 

Ex.  1445.  Find  the  volume  of  the  solid  generated  by  revolving  the 
triangle  of  Ex.  1444  about  side  20  as  an  axis. 

Ex.  1446.  Find  the  volume  of  the  solid  generated  by  revolving  the 
triangle  of  Ex.  1444  about  its  longest  side  as  an  axis. 

Ex.  1447.  The  slant  height  of  a  right  circular  cone  is  20  inches,  and 
the  circumference  of  its  base  4  IT  inches.  A  plane  parallel  to  the  base 
cuts  off  a  cone  whose  slant  height  is  8  inches.  Find  the  lateral  area  and 
the  volume  of  the  frustum  remaining. 


BOOK  VIII  415 

Ex.  1448.  A  cone  has  an  altitude  of  12.5  feet  and  a  base  whose 
radius  is  8.16  feet;  the  base  of  a  cylinder  having  the  same  volume  as  the 
cone  has  a  radius  of  6.25  feet.  Find  the  altitude  of  the  cylinder. 

Ex.  1449.  A  log  20  feet  long 
is  3  feet  in  diameter  at  the  top  end 
and  4  feet  in  diameter  at  the  butt 
end. 

(a)  How  many  cubic  feet  of 
wood  does  the  log  contain  ? 

(6)  How  many  cubic  feet  are  there  in  the  largest  piece  of  square 
timber  that  can  be  cut  from  the  log? 

(c)  How  many  cubic  feet  in  the  largest  piece  of  square  timber  the  same 
size  throughout  its  whole  length  ? 

(d)  How  many  board  feet  does  the  piece  of  timber  in  (c)  contain,  a 
board  foot  being  equivalent  to  a  board  1  foot  square  and  1  inch  thick  ? 

HIXT.  In  (6)  the  larger  end  is  square  ABCD.  What  is  the  smaller 
end  ?  In  (c)  one  end  is  square  EFGH.  What  is  the  other  end  ? 

Ex.  1450.  The  base  of  a  cone  has  a  radius  of  16  inches.  A  section 
of  the  cone  through  the  vertex,  through  the  center  of  the  base,  and  per- 
pendicular to  the  base,  is  a  triangle  two  of  whose  sides  are  20  inches 
and  24  inches,  respectively.  Find  the  volume  of  the  cone. 

Ex.  1451.  The  hypotenuse  of  a  right  triangle  is  10  inches  and  one 
side  8  inches  ;  find  the  area  of  the  surface  generated  by  revolving  the  tri- 
angle about  its  hypotenuse  as  an  axis. 

Ex.  1452.  A  tin  pail  in  the  form  of  a  frustum  of  a  righi  circular 
cone  is  8  inches  deep ;  the  diameters  of  its  bases  are  8£  inches  and  10* 
inches,  respectively.  How  many  gallons  of  water  will  it  hold  ?  (One 
liquid  gallon  contains  231  cubic  inches.) 

Ex.  1453.  The  altitude  of  a  cone  is  12  inches.  At  what  distances 
from  the  vertex  must  planes  be  passed  parallel  to  the  base  to  divide  the 
cone  into  four  equivalent  parts  ? 

HINT.     See  Ex.  1430. 

Ex.  1454.  Find  the  volume  of  the  solid  generated  by  an  equilateral 
triangle,  whose  side  is  a,  revolving  about  one  of  its  sides  as  an  axis. 

Ex.  1455.  Regular  hexagonal  prisms  are  inscribed  in  and  circum- 
scribed about  a  right  circular  cylinder.  Find  (a)  the  ratio  of  the  lateral 
areas  of  the  three  solids  ;  (6)  the  ratio  of  their  total  areas ;  (c)  the  ratio 
of  their  volumes. 


416  SOLID  GEOMETRY 

Ex.  1456.  How  many  miles  of  platinum  wire  ^  of  an  inch  in  diam- 
eter can  be  made  from  1  cubic  foot  of  platinum  ? 

Ex.  1457.  A  tank  in  the  form  of  a  right  circular  cylinder  is  5  feet 
long  and  the  radius  of  its  base  is  8  inches.  If  placed  so  that  its  axis  is 
horizontal  and  filled  with  gasoline  to  a  depth  of  12  inches,  how  many 
gallons  of  gasoline  will  it  contain  ? 

HINT.     See  Ex.  1024. 

Ex.  1458.  Find  the  weight  in  pounds  of  an  iron  pipe  10  feet  long,  if 
the  iron  is  £  inch  thick  and  the  outer  diameter  of  the  pipe  is  4  inches. 
(1  cubic  foot  of  bar  iron  weighs  7780  ounces.) 

Ex.  1459.  In  a  certain  right  circular  cone  whose  altitude  and  radius 
of  base  are  equal,  the  total  surface  and  the  volume  have  the  same  numeri- 
cal value.  Find  the  volume  of  the  cone. 

Ex.  1460.  Two  cones  of  revolution  lie  on  opposite  sides  of  a  common 
base.  Their  slant  heights  are  12  and  5,  respectively,  and  the  sum  of  their 
altitudes  is  13.  Find  the  radius  of  the  common  base. 

Ex.  1461.  The  radii  of  the  lower  and  upper  bases  of  a  frustum  of  a 
right  circular  cone  are  R  and  R',  respectively.  Show  that  the  area  of  a 

TT  C  K  -4-  7?'^2 

section  midway  between  them  is  -^ — "*"      ;  . 

Ex.  1462.  A  plane  parallel  to  the  base  of  a  right  circular  cone  leaves 
three  fourths  of  the  cone's  volume.  How  far  from  the  vertex  is  this 
plane  ?  How  far  from  the  vertex  is  the  plane  if  it  cuts  off  half  the 
volume  ?  Answer  the  same  questions  for  a  cylinder. 

Ex.  1463.  Is  every  cone  cut  from  a  right  circular  cone  by  a  plane  par- 
allel to  its  base  necessarily  similar  to  the  original  cone  ?  why  ?  How  is 
it  with  a  cylinder  ?  why  ? 

Ex.  1464.  Water  is  carried  from  a  spring  to  a  house,  a  distance  of 
f  mile,  in  a  cylindrical  pipe  whose  inside  diameter  is  2  inches.  How 
many  gallons  of  water  are  contained  in  the  pipe  ? 

Ex.  1465.  A  square  whose  side  is  6  inches  is  revolved  about  one  of 
its  diagonals  as  an  axis.  Find  the  surface  and  the  volume  of  the  solid 
generated.  Can  you  tind  the  volume  of  the  solid  generated  by  revolving 
a  cube  about  one  of  its  diagonals  as  an  axis  ? 

HINT.  Make  a  cube  of  convenient  size  from  pasteboard,  pass  a  hat- 
pin through  two  diagonally  opposite  vertices,  and  revolve  the  cube  rapidly. 

Ex.  1466.  Given  V  the  volume,  and  R  the  radius  of  the  base,  of  a 
right  circular  cylinder.  Find  the  lateral  area  and  total  area. 

Ex.  1467.  Given  the  total  area  2\  and  the  altitude  //,  of  a  right 
circular  cylinder.  Find  the  volume. 


BOOK   IX 


THE  SPHERE 

900.  Def.     A  sphere  is  a  solid  closed  figure  whose  boundary 
is  a  curved  surface  such  that  all  straight  lines  to  it  from  a 
fixed  poiot  within  are  equal. 

901.  Defs.     The  fixed  point  within 
the    sphere    is    called    its    center;     a 
straight    line    joining    the    center    to 
any  point  on  the  surface  is  a  radius ; 
a   straight  line   passing  through  the 
center  and  having  its  extremities  on 
the  surface  is  a  diameter. 

902.  From  the  above  definitions  and 
from  the  definition   of  equal  figures, 
§  18,  it  follows  that : 

(a)  All  radii  of  the  same  sphere,  or  of  equal  spheres,  are  equal. 

(b)  All  diameters  of  the  same  sphere,  or  of  equal  spheres,  are 
equal. 

(c)  Spheres  having  equal  radii,  or  equal  diameters,  are  equal. 

(d)  A  sphere  may  be  generated  by  the  revolution  of  a  circle 
about  a  diameter  as  an  axis. 


Ex.  1468.     Find  the  locus  of  all  points  that  are  3  inches  from  the  sur- 
face of  a  sphere  whose  radius  is  7  inches. 

Ex.  1469.     The  three  edges  of  a  trihedral  angle  pierce  the  surface  of  a 
sphere.     Find  the  locus  of  all  points  of  the  sphere  that  are  : 

(a)  Equidistant  from  the  three  edges  of  the  trihedral  angle, 

(6)  Equidistant  from  the  three  faces  of  the  trihedral  angle. 

Ex.  1470.     Find  a  point  in  a  plane  equidistant  from  three  given  points 
in  space. 

Ex.  1471.     Find  the  locus  of  all  points  in  space  equidistant  from  the 
three  sides  of  a  given  triangle. 

417 


418 


SOLID   GEOMETRY 


PROPOSITION  I.     THEOREM 

903.  Every  section  of  a  sphere  made  by  a  plane  is  a 
circle. 


Given   AMBN  a  section  of  sphere  0  made  by  a  plane. 
To  prove   section  AMBN  a  O. 


ARGUMENT 

1.  From  0  draw  OQ  _L  section  AMBN. 

2.  Join  Q  to  C  and  D,  any  two  points  on 

the  perimeter  of  section  AMBN.     Draw 
OC  and  OD. 

3.  Ill  rt.  A  OQC  and  OQD,  OQ  =  OQ. 

4.  OC  =  OD. 

5.  /.A  OQC  =  A  OQD. 

6.  /.  QC  =  QD-,  i.e.  any  two  points  on  the 

perimeter  of  section  AMBN  are  equi- 
distant from  Q. 
1.    .'.  section  AMBN  is  a  O.  Q.E.D. 

904.  Def.     A    great    circle    of    a 
sphere  is  a  section  made  by  a;  plane 
which  passes  through  the  center  of 
the  sphere,  as  O  CRDS. 

905.  Def.      A    small    circle    of    a 
sphere  is  a  section  made  by  a  plane 
which  does  not  pass   through  the 
center  of  the  sphere,  as  O  AMBN. 


REASONS 

1.  §  639. 

2.  §  54,  15. 


3.  By  iden. 

4.  §  902,  a. 

5.  §  211. 

6.  §  110. 


7.    §  276. 


BOOK  IX  419 

906.  Def.     The  axis  of  a  circle  of  a  sphere  is  the  diameter  of 
the  sphere  which  is  perpendicular  to  the  plane  of  the  circle. 

907.  Def.     The  poles  of  a  circle  of  a  sphere  are  the  extremi- 
ties of  the  axis  of  the  circle. 


Ex.  1472.  Considering  the  earth  as  a  sphere,  what  kind  of  circles 
are  the  parallels  of  latitude  ?  the  equator  ?  the  meridian  circles  ?  What 
is  the  axis  of  the  equator  ?  of  the  parallels  of  latitude  ?  What  are  the 
poles  of  the  equator  ?  of  the  parallels  of  latitude  ? 

Ex,  1473.  The  radius  of  a  sphere  is  17  inches.  Find  the  area  of  a 
section  made  by  a  plane  8  inches  from  the  center. 

Ex.  1474.  The  area  of  a  section  of  a  sphere  45  centimeters  from  the 
center  is  784  TT  square  centimeters.  Find  the  radius  of  the  sphere. 

Ex.  1475.  The  area  of  a  section  of  a  sphere  7  inches  from  the  center 
is  576  TT.  Find  the  area  of  a  section  8  inches  from  the  center. 


908.  The  following  are  some  of  the  properties  of  a  sphere ; 
the  student  should  prove  the  correctness  of  each : 

(a)  In  equal  spheres,  or  in  the  same  sphere,  if  two  sections  are 
equal,  they  are  equally  distant  from  the  center,  and  conversely. 

HINT.     Compare  with  §  307. 

(6)  In  equal  spheres,  or  in  the  same  sphere,  if  two  sections  are 
unequal,  the  greater  section  is  at  the  less  distance  from  the  center, 
and  conversely.  (HINT.  See  §§308, 310.) 

(c)  In  equal  spheres,  or  in  the  same,  sphere,  all  great  circles  are 
equal.      (HINT.     See  §  279,  c.) 

(d)  'TJie  axis  of  a  small  circle  of  a  sphere  passes  through  the 
center  of  the  circle,  and  conversely. 

(e)  Any  two  great  circles  of  a  sphere  bisect  each  other. 

(/)  Every  great  circle  of  a  sphere  bisects  the  surface  and  the 
sphere. 

(</)  Through  two  points  on  the  surface  of  a  sphere,  not  the  ex- 
tremities of  a  diameter,  there  exists  one  and  only  one  great  circle. 

(h)  Through  three  points  on  the  surface  of  a  sphere  there  exists 
one  and  only  one  circle 

909.  Def.      The  distance  between  two  points  on  the  surface 
of  a  sphere  is  the  length  of  the  minor  arc  of  the  great  circle 
joining  them. 


420  SOLID   GEOMETRY 

• 
PROPOSITION  II.     THEOREM 

910.  All  points  on  the  circumference  of  a  circle  of  a 
sphere  are  equidistant  from  either  pole  of  the  circle. 


Given  C  and  D  any  two  points  on  the  circumference,  and 
P  and  R  the  poles,  of  O  AMBN. 

To  prove  arc  PC=  arc  PD  and  arc  RG  =  arc  RD. 
The  proof  is  left  as  an  exercise  for  the  student. 
HINT.  Apply  §  298. 

911.  Def.     The    polar  distance  of  a  circle  of  a  sphere  is  the 
distance  between  any  point  on  its  circumference  and  the  nearer 
pole  of  the  circle. 

912.  Cor.  I.    The  polar  distance  of  a  great  circle  is  a 
quadrant. 

913.  Cor.  II.    In  equal  spheres,  or  in  the  same  sphere, 
the  polar  distances  of  equal  circles  are  equal. 

Ex.  1476.  What  is  the  locus  of  all  points  on  the  surface  of  the  earth 
at  a  quadrant's  distance  from  the  north  pole  ?  from  the  south  pole  ?  from 
the  equator  ?  from  a  point  P  on  the  equator?  at  a  distance  of  23|°  from 
the  south  pole  ?  2%%°  from  the  equator  ?  180°  from  the  north  pole  ? 

Ex.  1477.  Considering  the  earth  as  a  sphere  with  a  radius  of  4000 
miles,  calculate  in  miles  the  polar  distance  of  :  (a)  the  Arctic  Circle  ;  (6) 
the  Tropic  of  Cancer  ;  (c)  the  equator. 

Ex.  1478.  State  a  postulate  for  the  construction  of  a  circle  on  the 
surface  of  a  sphere  corresponding  to  §  122,  the  postulate  for  the  construc- 
tion of  a  circle  in  a  plane. 


BOOK   IX 
PROPOSITION  III.     THEOREM 


421 


914.  A  point  an  the  surface  of  a  sphere  at  the  distance  of 
a  quadrant  from  each  of  two  other  points  (not  the  extrem- 
ities of  the  same  diameter}  on  the  surface,  is  the  pole  of 
the  great  circle  passing  through  these  two  points. 


Given   PC  and  PD  quadrants  of  great  ©  of  sphere  0,  and 
ACDB  a  great  O  passing  through  points  C  and  D. 
To  prove   P  the  pole  of  great  O  ACDB. 

ARGUMENT 

Draw  0(7,  OD,  and  OP. 
PC  =  90°  and  PD  =  90°. 
.-.  A  POC  and  POD  are  rt.  A\  i.e.  OP  JL  OC 

and  OD 

.'.  OP  _L  the  plane  of  O  ACDB. 
.'.  OP  is  the  axis  of  O  ACDB. 
.'.  P  is  the  pole  of  great  O  ACDB.        Q.E.D. 


REASONS 

1. 

§ 

54,  15. 

2. 
3. 

By  hyp 

§358. 

4. 

§ 

622. 

5. 

§ 

906. 

6. 

§ 

907. 

Ex.  1479.  Assuming  the  chord  of  a  quadrant  of  a  great  circle  of  a 
sphere  to  be  given,  construct  with  compasses  an  arc  of  a  great  circle 
through  two  given  points  on  the  surface  of  the  sphere. 

Ex.  1480.  If  the  planes  of  two  great  circles  are  perpendicular  to  each 
other,  each  passes  through  the  poles  of  the  other. 

Ex.  1481.  Find  the  locus  of  all  points  in  space  equidistant  from  two 
given  points  and  at  a  given  distance  d  from  a  third  given  point. 

Ex.  1482.  Find  the  locus  of  all  points  in  space  at  a  distance  d  from 
a  given  point  and  at  a  distance  m  from  a  given  plane. 

Ex.  1483.  Find  the  locus  of  all  points  in  space  equidistant  from  two 
given  points  and  also  equidistant  from  two  given  parallel  lines. 


422 


SOLID   GEOMETRY 


PROPOSITION  IV.     THEOREM 

915.  The  intersection  of  two  spherical  surfaces  is  the 
circumference  of  a  circle. 


Given  two  spherical  surfaces  generated  by  intersecting  cir- 
cumferences O  and  Q  revolving  about  line  MN  as  an  axis. 

To  prove  the  intersection  of  the  two  spherical  surfaces  the 
circumference  of  a  Q. 

OUTLINE  OF  PROOF 

1.  Show  that  MN  _L  AB  at  its  mid-point  C  (§  328). 

2.  Show  that  AC,  revolving  about  axis  MN,  generates  a  plane. 

3.  Show  that  A  generates  the  circumference  of  a  O. 

4.  The  locus  of  A  is  the  intersection  of  what  (§  614)  ? 

5.  .'.  the  intersection  of  the  two  spherical  surfaces  is  the  cir- 
cumference of  a  O.  Q.E.D. 

Ex.  1484.  Find  the  locus  of  all  points  in  space  6  inches  from  a 
given  point  P  and  10  inches  from  another  given  point  Q. 

Ex.  1485.  The  radii  of  two  intersecting  spheres  are  12  inches  and 
16  inches,  respectively.  The  line  joining  their  centers  is  24  inches.  Find 
the  circumference  and  area  of  their  circle  of  intersection. 


916.  Def.    An  angle  formed  by  two  intersecting  arcs  of  circles 
is  the  angle  formed  by  tangents  to  the  two  arcs  at  their  point 
of     intersection ;    thus    the    Z.    formed 

by  arcs  AB  and  AC  is  plane  /_  DAE.  ^^  \ 

917.  Def.       A    spherical    angle    is    an 
angle  formed   by  two  intersecting  arcs 


of  great  circles  of  a  sphere.' 


D  C 


*  A  different  meaning  i8  sometimes  attached  to  the  expression  "  spherical  an^le  "  in 
advanced  mathematics. 


BOOK  IX  423 

. 
PROPOSITION  V.     THEOREM 

918.  Ji  spherical  angle  is  measured  by  the  arc  of  a 
great  circle  having  the  vertex  of  the  angle  as  a  pole  and 
intercepted  by  the  sides  of  the  angle,  prolonged  if  nec- 
essary. P 


Given  spherical  Z  APB,  with  CD  an  arc  of  a  great  O  whose  pole 
is  P  and  which  is  intercepted  by  sides  PA  and  PB  of  Z  APB. 
To  prove    that  Z  APB  oc  CD. 

OUTLINE  OF  PROOF 

1.  Draw  radii  OP,  0(7,  and  OD. 

2.  From  P  draw  PR  tangent  to  PA  and  PT  tangent  to  PB. 

3.  Prove  OC  and  OD  each  J_  OP. 

4.  Prove  OC  II  PR^OD  II  PT,  and  hence  Z  COD  =^Z  RPT. 

5.  But  Z  <70£  oc  SB;  .'.  Z  flPT,  i.g.  Z  ^P5  cc  cz).  Q.E.D. 

919.  Cor,  I.    A  spherical  angle  is  equal  to  the  plane 
angle  of  the  dihedral  angle  formed  by  the  planes  of  the 
sides  of  the  angle. 

920.  Cor.  II.    The  sum  of  all  the  spherical  angles  about 
a  point  on  the  surface  of  a  sphere  equals  four  right  angles. 


Ex.  1486.  By  comparison  with  the  definitions  of  the  corresponding 
terms  in  plane  geometry,  frame  exact  definitions  of  the  following  classes 
of  spherical  angles  :  acute,  right,  obtuse,  adjacent,  complementary,  supple- 
mentary, vertical. 

Ex.  1487.     Any  two  vertical  spherical  angles  are  equal. 

Ex.  1488.  If  one  great  circle  passes  through  the  pole  of  another 
great  circle,  the  circles  are  perpendicular  to  each  other. 


424 


SOLID   GEOMETRY 


LINES   AND   PLANES   TANGENT   TO   A  SPHERE 

921.  Def .     A   straight  line  or  a  plane  is  tangent  to  a  sphere 

if,  however  far  extended,  it  meets  the  sphere  in  one  and  only 
one  point. 

922.  Def.     Two  spheres  are  tangent  to  each  other  if  they 
have  one  and  only  one  point  in  common.     They  are  tangent 
internally  if  one  sphere  lies  within  the  other,  and  externally  if 
neither  sphere  lies  within  the  other. 

PROPOSITION  VI.     THEOREM 

923.  A  plane  tangent  to  a  sphere  is  perpendicular  to 
the  radius  drawn  to  the  point  of  tangency. 


Given  plane  AB  tangent  to  sphere  0  at  T,  and  OT  a  radius 
drawn  to  the  point  of  tangency. 
To  prove  plane  AB  JL  OT. 
The  proof  is  left  as  an  exercise  for  the  student. 

924.  Question.     What  changes  are  necessary  in  the  proof  of  §  313 
to  make  it  the  proof  of  §  923  ? 

925.  Cor.  I.    (Converse  of  Prop.    VI).    A  plane  perpen- 
dicular to  a  radius  of  a  sphere  at  its  outer  extremity  is 
tangent  to  the  sphere. 

HINT.     See  §  314.  

Ex.  1489.     A  straight  line  tangent  to  a  sphere  is  perpendicular  to  the 
radius  drawn  to  the  point  of  tangency. 

Ex.  1490.    State  and  prove  the  converse  of  Ex.  1489. 


BOOK  IX  425 

Ex.  1491.  Two  lines  tangent  to  a  sphere  at  the  same  point  determine 
a  plane  tangent  to  the  sphere  at  that  point. 

Ex.  1492.  Given  a  point  P  on  the  surface  of  sphere  O.  Explain  how 
to  construct :  (a)  a  line  tangent  to  sphere  0  at  P ;  (6)  a  plane  tangent 
to  sphere  O  at  P. 

Ex.  1493.  Given  a  point  H  outside  of  sphere  Q.  Explain  how  to  con- 
struct :  (a)  a  line  through  H  tangent  to  sphere  Q  ;  (£>)  a  plane  through  R 
tangent  to  sphere  Q. 

HINT.     Compare  with  §  373. 

Ex.  1494.  Two  planes  tangent  to  a  sphere  at  the  extremities  of  a 
diameter  are  parallel. 

Ex.  1495.  If  the  straight  line  joining  the  centers  of  two  spheres  is 
equal  to  the  sum  of  their  radii,  the  spheres  are  tangent  to  each  other. 

HINT.  Show  that  the  radius  of  the  O  of  intersection  of  the  two 
spheres  (§  915)  is  zero. 

926.  Def.     A  polyhedron  is  circumscribed  about  a  sphere  if 

each  face  of  the  polyhedron  is  tangent  to  the  sphere. 

927.  Def.     If  a  polyhedron  is  circumscribed  about  a  sphere, 
the  sphere  is  said  to  be  inscribed  in  the  polyhedron. 

928.  Def.     A  polyhedron  is  inscribed  in  a  sphere  if  all  its 

vertices  are  on  the  surface  of  the  sphere. 

929.  Def.     If  a  polyhedron  is  inscribed  in  a  sphere,  the 
sphere  is  said  to  be  circumscribed  about  the  polyhedron. 


Ex.  1496.  Find  the  edge  of  a  cube  inscribed  in  a  sphere  whose  radius 
is  10  inches. 

Ex.  1497.  Find  the  volume  of  a  cube :  (a)  circumscribed  about  a 
sphere  whose  radius  is  8  inches  ;  (6)  inscribed  in  a  sphere  whose  radius 
is  8  inches. 

Ex.  1498.  A  right  circular  cylinder  whose  altitude  is  8  inches  is  in- 
icribed  in  a  sphere  whose  radius  is  6  inches.  Find  the  volume  of  the 
cylinder. 

Ex.  1499.  A  right  circular  cone,  the  radius  of  whose  base  is  8  inches, 
is  inscribed  in  a  sphere  with  radius  12  inches.  Find  the  volume  of  the 
cone. 

Ex.  1500.  Find  the  volume  of  a  right  circular  cone  circumscribed 
about  a  regular  tetrahedron  whose  edge  is  «. 


426 


SOLID   GEOMETRY 


PROPOSITION  VII.     PROBLEM 

930.   To  inscribe  a  sphere  in  a  given  tetrahedron, 

V 


D 


Given   tetrahedron  V-ABC. 

To  inscribe   a  sphere  in  tetrahedron  V-ABC. 

I.    Construction 

1.  Construct  planes  RABS,  SBCT,  and  TCAR  bisecting  dihedral 
A  whose  edges  are  AB,  BC,  and  CA,  respectively.     §  691. 

2.  From  0,  the  point  of  intersection  of  the   three   planes, 
construct  OF  A.  plane  ABC.     §  637. 

3.  The  sphere  constructed  with  0  as  center  and  OF  as  radius 
will  be  inscribed  in  tetrahedron  V-ABC. 


II.    Proof 

ARGUMENT 

1.  Plane   RABS,  the  bisector  of  dihedral 

Z.  AB,  lies  between  planes  ABV  and 
ABC;  i.e.  it  intersects  edge  VC  in 
some  point  as  D. 

2.  .-.  plane  RABS  intersects  plane  BCV  in 

line  BD  and  plane  ACV  in  line  AD. 

3.  Plane  SBCT  lies  between  planes  BCV 

and  ABC;  i.e.  it  intersects  plane  RASB 
in  a  line  through  B  between  BA  and 
BD,  as  £& 

4.  Similarly  plane  TCAR  intersects  plane 

RABS  iii  a  line  through  A  between 


REASONS 
1.    By  cons. 


2.  §  616. 

3.  By  cons. 


4.  By  steps  sim- 
ilar to  1    .",. 


BOOK   IX 


427 


ARGUMENT 

AB  and  AD,  as  AR ;  and  plane  SECT 
intersects  plane  TCAR  in  a  line 
through  C  as  CT. 

5.  But  AR  and  BS  pass  through  the  in- 

terior of  A  ABD. 

6.  .-.  AR  and  BS  intersect  in  some  point 

as  0,  within  A  ABD. 

1.    .-.  AR,  BS,  and   CT  are  concurrent  in 
point  0. 

8.  From  0  draw  OH,  OK,  and  0£  _L  planes 

VAB,  VBC,  and  FC4,  respectively. 

9.  v  0  is  in  plane  OAB,  OF  =  OH. 

10.  v  0  is  in  plane  OBC,  OF  =  OK. 

11.  v  0  is  in  plane  OCA,  OF  =  OL. 

12.  .-.  OF=  OH  =  OK=  OL. 

13.  .-.  each  of  the  four  faces  of  the  tetra- 

hedron is  tangent  to  sphere  0. 

14.  .-.  sphere  O  is  inscribed  in  tetrahedron 

V-ABC.  Q.E.D. 


REASONS 

5.  Args.  3  &  4. 

6.  §  194. 

7.  §  617,  I. 

8.  §  639. 

9.  §  688. 

10.  §  688. 

11.  §  688. 

12.  §  54,  1. 

13.  §  925. 

14.  §§  926,  927. 


III.    Discussion 

The  discussion  is  left  as  an  exercise  for  the  student. 


Ex.  1501.  The  six  planes  bisecting  the  dihedral  angles  of  a  tetrahe- 
dron meet  in  a  point  which  is  equidistant  from  the  four  faces  of  the 
tetrahedron. 

HINT.     Compare  with  §  258. 

Ex.  1502.     Inscribe  a  sphere  in  a  given  cube. 

Ex.  1503.  The  volume  of  any  tetrahedron  is  equal  to  the  product  of 
its  surface  and  one  third  the  radius  of  the  inscribed  sphere. 

HINT.     See  §§  491  and  492. 

Ex.  1504.  Find  a  point  within  a  triangular  pyramid  such  that  the 
planes  determined  by  the  lines  joining  this  point  to  the  vertices  shall  di- 
vide  the  pyramid  into  four  equivalent  parts. 

HINT.     Compare  with  Ex.  1094. 


428 


SOLID   GEOMETRY 


PROPOSITION  VIII.     PROBLEM 


931.   To  circumscribe  a  sphere  about  a  given  tetrahe- 
dron. 


Given   tetrahedron  V-ABC. 

To  circumscribe   a  sphere  about  tetrahedron  V-ABC. 

I.    Construction 

1.  Through  D,  the  mid-point  of  AB,  construct  plane  DO_L  AB ; 
through  E9  the  mid-point  of  BC,  construct  plane  EOA.BC;  and 
through  F,  the  mid-point  of  VB,  construct  plane  FO  _L  VB. 

2.  Join  0,  the  point  of  intersection  of  the  three  planes,  to  A. 

3.  The  sphere  constructed  with  0  as  center  and  OA  as  radius 
will  be  circumscribed  about  tetrahedron  V-ABC. 

II.    Outline  of  Proof 

1.  Prove  that  the  three  planes  OD,  OE,  and  OF  intersect  each 
other  in  three  lines. 

2.  Prove  that  these  three  lines  of  intersection  meet  in  a 
point,  as  0. 

3.  Prove  that  OA  =  OB  =  OC  =  OF. 

4.  .  •.  sphere  0  is  circumscribed  about  tetrahedron  V-ABC. 

Q.E.D. 

932.  Cor.     Four  points  not  in  the  same  plane  determine 
a  sphere. 

933.  Questions.    Are  the  methods  used  in  §§  930  and  931  similar 
to  those  used  in  §§  321  and  323  ?    In  §  930  could  the  lines  forming  the 
edges  of  the  dihedral  angles  bisected  be  three  lines  meeting  in  one  vertex  ? 
In  §  931  could  the  three  edges  bisected  be  three  lines  lying  in  the  same  face  ? 


BOOK  IX  429 

Ex.  1505.  The  six  planes  perpendicular  to  the  edges  of  a  tetrahedron 
at  their  mid-points  meet  in  a  point  which  is  equidistant  from  the  four  ver- 
tices of  the  tetrahedron. 

Ex.  1506.  The  four  lines  perpendicular  to  the  faces  of  a  tetrahedron, 
and  erected  at  their  centers,  meet  in  a  point  which  is  equidistant  from  the 
four  vertices  of  the  tetrahedron. 

Ex.   1507.     Circumscribe  a  sphere  about  a  given  cube. 

Ex.  1508.  Circumscribe  a  sphere  about  a  given  rectangular  paral- 
lelepiped. Can  a  sphere  be  inscribed  in  any  rectangular  parallelepiped  ? 
Explain. 

Ex.  1509.  Find  a  point  equidistant  from  four  points  in  space  not  all 
in  the  same  plane. 

SPHERICAL  POLYGONS 

934.  Def.     A  line  on  the  surface  of  a  sphere  is  said  to  be 
closed  if  it  separates  a  portion  of  the  surface  from  the  remain- 
ing portion. 

935.  Def.     A  closed  figure  on  the  surface  of  a  sphere  is  a 
figure  composed  of  a  portion  of  the  surface  of  the  sphere  and 
its  bounding  line  or  lines. 

936.  Defs.     A  spherical  polygon  is  a  closed  figure  on  the 
surface  of  a  sphere  whose  boundary  is  composed  of  three  or 
more  arcs  of  great  circles,  as  ABCD. 


The  bounding  arcs  are  called  the  sides  of  the  polygon,  the 
points  of  intersection  of  the  arcs  are  the  vertices  of  the  poly- 
gon, and  the  spherical  angles  formed  by  the  sides  are  the 
angles  of  the  polygon. 


430  SOLID   GEOMETRY 

937.  Def.     A  diagonal  of  a  spherical  polygon  is  an  arc  of  a 

great  circle  joining  any  two  non-adjacent  vertices. 

938.  Def.     A  spherical  triangle  is  a  spherical  polygon  having 
three  sides. 


Ex.  1510.  By  comparison  with  the  definitions  of  the  corresponding 
terms  in  plane  geometry,  frame  exact  definitions  of  the  following  classes 
of  spherical  triangles  :  scalene,  isosceles,  equilateral,  acute,  right,  obtuse, 
and  equiangular. 

Ex,  1511.  With  a  given  arc  as  one  side,  construct  an  equilateral 
spherical  triangle. 

HINT.     Compare  the  cons.,  step  by  step,  with  §  124. 

Ex.  1512.  With  three  given  arcs  as  sides,  construct  a  scalene  spheri- 
cal triangle. 

939.  Since  each  side  of  a  spherical  polygon  is  an  arc  of  a 
great  circle  (§  936),  the  planes  of  these  arcs  meet  at  the  center 
of  the  sphere  and  form  at  that  point  a  polyhedral  angle,  as 
polyhedral  Z  0-ABCD. 

This  polyhedral  angle  and  the  spheri- 
cal polygon  are  very  closely  related. 
The  following  are  some  of  the  more 
important  relations;  the  student 
should  prove  the  correctness  of  each : 

940.  (a)   The   sides   of   a    spherical 
polygon  have  the  same  measures  as  the 
corresponding  face  angles  of  the  poly- 
hedral angle. 

(b)  The  angles  of  a  spherical  polygon  have  the  same  measures 
as  the  corresponding  dihedral  angles  of  the  polyhedral  angle. 

Thus,  sides  AB,  BC,  etc.,  of  spherical  polygon  ABGD  have 
the  same  measures  as  face  AAOB,  BOG,  etc.,  of  polyhedral 
Z  0-ABCD ;  and  spherical  A  ABC,  BCD,  etc.,  have  the  same 
measures  as  the  dihedral  A  whose  edges  are  OB,  OC,  etc. 

These  relations  make  it  possible  to  establish  certain  prop- 
erties of  spherical  polygons  from  the  corresponding  known 
properties  of  the  polyhedral  angle,  as  in  §  §  941  and  942. 


BOOK  IX 
PROPOSITION  IX.     THEOREM 


431 


941.   The  sum  of  any  two  sides  of  a  spherical  triangle 
is  greater  than  the  third  side. 


Given    spherical  A  ABC. 
To  prove    AB  +  BC  >  CA. 

ARGUMENT 

1.  Z  AOB  4-  Z  50C  >  Z  COA. 

2.  Z  ^1O.B  QC  AB,  Z  BOC&BC,  Z  COJ  QC 


3.    .'.  AB  -f 


> 


Q.E.D. 


REASONS 

1.  §710. 

2.  §940,  a, 

3.  §  362,  6. 


PROPOSITION  X.     THEOREM 

942.   The  sum  of  the  sides  of  any  spherical  polygon  is 
less  than  360°. 


Given   spherical  polygon  ABC  •••  with  n  sides, 

To  prove    AB  +  BC  +  •••  <  360°. 
HINT.   See  §§  712  and  940,  a. 


432 


SOLID  GEOMETRY 


Ex.  1513.  In  spherical  triangle  ABC,  arc  AB  =  40°  and  arc  BC  = 
80°.  Between  what  limits  must  arc  CA  lie  ? 

Ex.  1514.  Any  side  of  a  spherical  polygon  is  less  than  the  sum  of  the 
remaining  sides. 

Ex.  1515.  If  arc  AB  is  the  perpendicular  bisector  of  arc  CD,  every 
point  on  the  surface  of  the  sphere  and  not  in  arc  AB  is  unequally  distant 
from  C  and  D. 

Ex.  1516.  In  a  spherical  quadrilateral,  between  what  limits  must  the 
fourth  side  lie  if  three  sides  are  60°,  70°,  and  80°?  if  three  sides  are  40°, 
60°,  and  70°  ? 

Ex.  1517.   Any  side  of  a  spherical  polygon  is  less  than  180°. 

HINT.     See  Ex.  1514.        

943.  If,  with  A,  B,  and  (7  the  vertices  of  any  spherical  tri- 
angle as  poles,  three  great  circles  are  constructed,  as  B'c'ED, 


C'A'DJ  and  EA'B',  the  surface  of  the  sphere  will  be  divided  into 
eight  spherical  triangles,  four  of  which  are  seen  on  the  hemi- 
sphere represented  in  the  figure.  Of  these  eight  spherical  tri- 
angles, A'B'C'  is  the  one  and  only  one  that  is  so  situated  that  A 
and  A'  lie  on  the  same  side  of  BC,  B  and  B1  on  the  same  side  of 
AC,  and  C  and  C'  on  the  same  side  of  AB.  This  particular  tri- 
angle A'B'C'  is  called  the  polar  triangle  of  triangle  ABC. 

944.  Questions.     In  the  figure  above,  A  A'B'C',  the  polar  of  &ABC, 
is  entirely  outside  of  A  ABC.     Can  the  two  &  be  so  constructed  that  : 
(a)  A'B'C1  is  entirely  within  ABC  ? 
(6)  A'B'C'  is  partly  outside  of  and  partly  within  ABC? 


Ex.  1518.     What  is  the  polar  triangle  of  a  spherical  triangle  all  of 
whose  sides  are  quadrants  ? 


,  BOOK   IX 


PROPOSITION  XI.     THEOREM 


433 


945.  If  one  spherical  triangle  is  the  polar  of  another, 
then  the  second  is  the  polar  of  the  first. 


Given  A  A'B'C'  the  polar  of  A  ABC. 
To  prove   A  ABC  the  polar  of  A  A'B 


1.  A  is  the  pole  of  B'c' ;  i.e.  AC'  is  a  quad- 

rant. 

2.  B  is  the  pole  of  A'C' ;  i.e.  BC1  is  a  quad- 

rant. 

3.  /.  C'  is  the  pole  of  AB. 

4.  Likewise  B'  is  the  pole  of  AC,  and  A'  is 

the  pole  of  BC. 

5.  /.A  ABC  is  the  polar  of  A  A'B'C'.  Q.E.D. 


REASONS 


1.    §  912. 


2.    §  912. 


3.  §914. 

4.  By  steps  simi- 

lar to  1-  3. 

5.  §  943. 


946.  Historical  Note.  The  properties  of  polar  triangles  were  dis- 
covered about  1626  A.D.  by  Albert  Girard,  a  Dutch  mathematician,  born 
in  Lorraine  about  1595.  They  were  also  discovered  independently  and 
about  the  same  time  by  Snell,  an  "infant  prodigy,"  who  at  the  age  of 
twelve  was  familiar  with  the  standard  mathematical  works  of  that  time 
and  who  is  remembered  as  the  discoverer  of  the  well-known  law  of 
refraction  of  light. 

Ex.  1519.  Determine  the  polar  triangle  of  a  spherical  triangle  having 
two  of  its  sides  quadrants  and  the  third  side  equal  to  70° ;  110° ;  (90  -  a)° ; 
(90  +  a)°. 


434 


SOLID   GEOMETRY 
PROPOSITION  XII.     THEOREM 


947.  In  two  polar  triangles  each  angle  of  one  and 
that  side  of  the  other  of  which  its  vertex  is  the  pole  are 
together  equal,  numerically,  to  180n- 

A 


Given-  polar  A  ABC  and  A'B'c',  with  sides  denoted  by  a,  6,  c, 
and  a',  &',  c',  respectively. 

To  prove:   (a)  Z  ^+a'  =  180°,  Z  £+6'  =  180°,  Z  <7+c'  =  lSO°; 
(6)  Z^'+a=180°,  Zj5'+&=1800,  Z  <7'+c=180°. 

(a)  ARGUMENT  ONLY 

1.  Let  arcs  AB  and  AC  (prolonged  if  necessary)  intersect  arc 
B'C'  at  D  and  E,  respectively ;  then  c'D  =  90°  and  EB'  =  90°. 

2.  .-.  C'D  +  EB'  =  180°. 

3.  . ' .     <fE  +  ED  +  ED  +  DB'  =  180° ;  i .e.  £7)  +  a'  -  180°. 

4.  But  ^D  is  the  measure  of  Z  4. 

5.  .\Z^  +  a'  =  180°. 

6.  Likewise    Z  5  +  6' =  180°,  and  Z  C  +  c'  =  180°.        Q.E.D. 
(6)  The  proof  of  (6)  is  left  as  an  exercise  for  the  student. 
HINT.     Let  BC  prolonged  meet  A'B'  at  H  and  A'C'  at  K. 

948.   Question.     In  the  history  of  mathematics,  why  are  polar  tri- 
angles frequently  spoken  of  as  supplemental  triangles  ? 


Ex.  1520.  The  angles  of  a  spherical  triangle  are  75°,  85°,  and  146°. 
Find  the  sides  of  its  polar  triangle. 

Ex.  1521.  If  a  spherical  triangle  is  equilateral,  its  polar  triangle  is 
equiangular ;  and  conversely. 


BOOK  IX 


435 


PROPOSITION  XIII.     THEOKEM 

949.    The  sum  of  the  angles  of  a  spherical  triangle  is 
greater  than  180°  and  less  than  540°. 

A 


, 


B 


ARGUMENT 


c          « 

Given   spherical  A  ABC  with  sides  denoted  by  a,  6,  and  c. 
To  prove   Z^  +  Z^  +  Zo  180°  and  <  540°. 


REASONS 

1.  §943. 

2.  §  947. 

3.  §  54,  2. 

4.  §  942. 

5.  §  54,  6. 

6.  §  938. 

7.  §  54,  6. 


1.  Let  A  A'B'C',  with  sides  denoted  by  a', 

b',  and  c',  be  the  polar  of  A  ABC. 

2.  Then  Z  ^  +  a'  =  180°,  Z  B  +  b'  =  180°, 


3.  . 

4.  But         a'  +  6'  +  c'  <  360°. 

5.  .-.  Z^  +  Z^+Z  c>180°. 

6.  Again,     a'  -f  6'  +  c'  >  0°. 
•7.  .-.  Z'4  +  Z£  +  Z  C<540°. 


Q.E.D. 


950.  Cor.     In  a  spherical  triangle  there  can  be   one, 
two,  or  three  right  angles ;  there  can  be  one,  two,  or  three 
obtuse  angles. 

951.  Note.     Throughout  the  Solid  Geometry  the  student's  attention 
has  constantly  been  called  to  the  relations  between  definitions  and  theorems 
of  solid  geometry  and  the  corresponding  definitions  and  theorems  of  plane 
geometry.     In  the  remaining  portion  of  the  geometry  of  the  sphere  there 
will  likewise  be  many  of  these  comparisons,  but  here  the  student  must  be 
particularly  on  his  guard  for  contrasts  as  well  as  comparisons.    For  ex- 


436 


SOLID   GEOMETRY 


ample,  while  the  sum  of  the  angles  of  a  plane  triangle  is  equal  to  exactly 
180°,  he  has  learned  (§  949)  that  the  sum  of  the  angles  of  a  spherical  tri- 
angle may  be  any  number  from  180°  to  540°  ;  while  a  plane  triangle  can 
have  but  one  right  or  one  obtuse  angle,  a  spherical  triangle  may  have  one, 
two,  or  three  right  angles  or  one,  two,  or  three  obtuse  angles  (§  950). 

If  the  student  will  recall  that,  considering  the  earth  as  a  sphere,  the 
north  and  south  poles  of  the  earth  are  the  poles  of  the  equator,  and  that 
all  meridian  circles  are.  great  circles  perpendicular  to  the  equator,  it  will 
make  his  thinking  about  spherical  triangles  more  definite. 

952.  Def.  A  spherical  triangle  containing  two  right  angles 
is  called  a  birectangular  spherical  triangle. 


953.  Def.  A  spherical  triangle  having  all  of  its  angles  right 
angles  is  called  a  trirectangular  spherical  triangle. 

Thus  two  meridians,  as  NA  and  NB,  making  at  the  north  pole 
an  acute  or  an  obtuse  Z,  form  with  the  equator  a  birectangular 
spherical  A.  If  the  Z  between  NA  and  NB  is  made  a  rt.  Z, 
A  ANB  becomes  a  trirectangular  spherical  A. 

Ex.  1522.  What  kind  of  arcs  are  NA  and  NB  ?  Then  what  arc 
measures  spherical  angle  ANB  ?  Are  two  sides  of  any  birectangular 
spherical  triangle  quadrants?  What  is  each  side  of  a  trirectangular 
spherical  triangle  ? 

Ex.  1523.  If  two  sides  of  a  spherical  triangle  are  quadrants,  the 
triangle  is  birectangular.  (HINT.  Apply  §  947.) 

Ex.  1524.  What  is  the  polar  triangle  of  a  trirectangular  spherical 
triangle  ? 

Ex.  1525.  An  exterior  angle  of  a  spherical  triangle  is  less  than  the 
sum  of  the  two  remote  interior  angles.  Compare  this  exercise  with  §  2ir>, 
Make  this  new  fact  clear  by  applying  it  to  a  birortangular  spherical 
triangle  whose  third  angle  is  :  (a)  acute;  (6)  right;  (c)  obtuse. 


BOOK  IX  487 

PROPOSITION  XIV.     THEOREM 

954.  In  equal  spheres,  or  in  the  same  sphere,  two  spher- 
ical triangles  are  equal : 

I.  If  a  side  and  the  two  adjacent  angles  of  one  are 
equal  respectively  to  a  side  and  the  two  adjacent  angles 
of  the  other; 

II.  If  two  sides  and  the  included  angle  of  one  are 
equal  respectively  to  two  sides  and  the  included  angle 
of  the  other ; 

III.  //  the  three  sides  of  one  are  equal  respectively  to 
the  three  sides  of  the  other  : 

provided  the  equal  parts  are  arranged  in  the  same  order. 
The  proofs  are  left  as  exercises  for  the  student. 
HINT.     In  each  of  the  above  cases  prove  the  corresponding  trihedral  A 
equal  (§§  702,  704)  ;  and  thus  show  that  the  spherical  &  are  equal. 

955.  Questions.     Compare  Prop.  XIV,  I  and  II,  with  §  702,  I  and 
II,  and  with  §§  105  and  107.     Could  the  methods  used  there  be  employed 
in  §  954  ?     Is  the  method  here  suggested  preferable  ?  why  ? 

956.  Def.     Two  spherical  polygons   are  symmetrical  if  the 

corresponding  polyhedral  angles  are  symmetrical. 


957.  The  following  are  some  of  the  properties  of  symmet- 
rical spherical  triangles ;  the  student  should  prove  the  correct- 
ness of  each : 

(a)  Symmetrical  spherical  triangles  have  their  parts  respectively 
equal,  but  arranged  in  reverse  order. 

(6)   Two  isosceles  symmetrical  spherical  triangles  are  equal. 

HINT.  Prove  (6)  by  superposition  or  by  showing  that  the  correspond- 
ing trihedral  A  are  equal. 


438 


SOLID   GEOMETRY 


PROPOSITION  XV.     THEOREM 

958.   In  equal  spheres,  or  in  the  same  sphere,  two  sym- 
metrical spherical  triangles  are  equivalent. 


A  A' 

Given   symmetrical   spherical  A  ABC  and   A'B'C'   in  equal 
spheres  0  and  O1. 

To  prove  spherical  AABC^  spherical  A  A'B'C'. 


ARGUMENT 

1.  Let  P  and  P1  be  poles  of  small  CD  through 

A,  B,  C,  and  A1,  B',  C1,  respectively. 

2.  Arcs  AB,  BC,  CA  are  equal,  respectively, 

to  arcs  A'B',  B'c',  C'A'. 

3.  .-.  chords  AB,  BC,  CA,  are  equal,  respec- 

tively, to  chords  A'B',  B'C',  C'A'. 

4.  .-.  plane  A  ABC  =  plane  A  A'B'C*. 

5.  .-.   O  ABC  =  O  A' B'C'. 

6.  Draw  arcs  of  great  ©  PA,  PB,  PC,  P'A', 

P'B',  and  P'C1. 
1.   Then 

PA  =  PB  =  PC  =  P'A'=  P'B'=  P'C'. 

8.  .-.  isosceles  spherical  A  APB  and  A' P'B' 

are  symmetrical. 

9.  .-.  A  APB  =  AA'P'B1. 

10.  Likewise  A  BPC  =  A  tf'P'C'  and 

A  CPA  =  A  C7'P'J'. 

11.  .-.  spherical  A 4£C=D=  spherical  AA'B'C*. 

Q.E.D. 


REASONS 

1.  §  908,  h. 

2.  §  957,  a. 

3.  §  298,  IL 

4.  §116. 

5.  §324. 

6.  §  908,  g. 

7.  §913. 


8.  §956. 

9.  §957,6. 

10.  By  steps  simi- 

lar to  8-9. 

11.  §  54,  2. 


BOOK  IX  439 

PROPOSITION  XVI.     THEOREM 

959.   In  equal  spheres,  or  in  the  same  sphere,  two  spheri- 
cal triangles  are  symmetrical,  and  therefore  equivalent  : 

I.  If  a  side  and  the  two  adjacent  angles  of  one  are 
equal  respectively  to  a  side  and  the  two  adjacent  angles 
of  the  other ; 

II.  If  two  sides  and  the  included  angle  of  one  are 
equal  respectively  to  two  sides  and  the  included  angle 
of  the  other; 

III.  If  the  three  sides  of  one  are  equal  respectively 
to  the  three  sides  of  the  other : 

provided  the  equal  parts  are  arranged  in  reverse  order. 
The  proofs  are  left  as  exercises  for  the  student. 
HINT.     In  each  of  the  above  cases  prove  the  corresponding  trihedral  A 
symmetrical  (§  709)  ;  and  thus  show  that  the  spherical  &  are  symmetrical. 


Ex.  1526.  The  bisector  of  the  angle  at  the  vertex  of  an  isosceles 
spherical  triangle  is  perpendicular  to  the  base  and  bisects  it. 

Ex.  1527.  The  arc  drawn  from  the  vertex  of  an  isosceles  spherical 
triangle  to  the  mid-point  of  the  base  bisects  the  vertex  angle  and  is  per- 
pendicular to  the  base. 

Ex.  1528.  State  and  prove  the  theorems  on  the  sphere  correspond- 
ing to  the  following  theorems  on  the  plane  : 

(1)  Every  point  in  the  perpendicular  bisector  of  a  line  is  equidistant 
from  the  ends  of  that  line  (§134). 

(2)  Every  point  equidistant  from  the  ends  of  a  line  lies  in  the  perpen- 
dicular bisector  of  that  line  (§  139). 

(3)  Every  point  in  the  bisector  of  an  angle  is  equidistant  from  the 
sides  of  the  angle  (§253). 

HINT.  In  the  figure  for  (3),  corresponding  to  the  figure  of  §  252,  draw 
PD  JL  AS  and  lay  off  BE  =  BD. 

Ex.  1529.  The  diagonals  of  an  equilateral  spherical  quadrilateral  are 
perpendicular  to  each  other.  Prove.  State  the  theorem  in  plane  geome- 
try that  corresponds  to  this  exercise. 

Ex.  1530.  In  equal  spheres,  or  in  the  same  sphere,  if  two  spherical 
triangles  are  mutually  equilateral,  their  polar  triangles  are  mutually  equi- 
angular ;  and  conversely. 


440 


SOLID   GEOMETRY 


PROPOSITION  XVII.     THEOREM 

960.  In  equal  spheres,  or  in  the  same  sphere,  if  two 
spherical  triangles  are  mutually  equiangular,  they  are 
mutually  equilateral,  and  are  either  equal  or  symmet- 
rical. 


\P 


Given   spherical  A  T  and  T1  in  equal  spheres,  or  in  the  same 
sphere,  and  mutually  equiangular. 

To  prove    T  and  T'  mutually  equilateral  and  either  equal  or 
symmetrical. 

ARGUMENT  REASONS 


1.  Let  A  P  and  P'  be  the  polars  of  A  T 

and  T1,  respectively. 

2.  T  and  T'  are  mutually  equiangular. 

3.  Then  P    and   P'    are    mutually    equi- 

lateral. 

4.  .*.  P  and  P'  are   either   equal  or  sym- 

metrical,   and    hence    are    mutually 
equiangular. 

5.  .*.  T  and  T1  are  mutually  equilateral. 

6.  .-.  T  and  T1  are   either  equal   or  sym- 


metrical. 


Q.E.D. 


1.    §  943. 


2.  By  hyp. 

3.  §  947. 

4.  §§    954,     III, 

and  959,  III. 

5.  §  947. 

6.  Same     reason 

as  4. 


Ex.  1531.     In  plane  geometry,  if  two  triangles  are  mutually  eqv 
angular,  what  can  be  said  of  them  ?    Are  they  equal  ?  equivalent  ? 

Ex.  1532.  Find  the  locus  of  all  points  of  a  sphere  that  are  equidis- 
tant from  two  given  points  on  the  surface  of  the  sphere  ;  from  two  givei 
points  in  space,  not  on  the  surface  of  the  sphere. 


BOOK  IX 


441 


PROPOSITION  XVIII.     THEOREM 

961.   The  base  angles  of  an  isosceles  spherical  triangle 
are  equal. 


Given  isosceles  spherical  A  ABC,  with  side  AS  =  side  BC. 
To  prove    Z.  A  =  Z  C. 
HINT.     Compare  with  §  111. 

PROPOSITION  XIX.     THEOREM 

962.   If  two  angles  of  a  spherical  triangle  are  equal, 
the  sides  opposite  are  equal, 


T    \ 


Given   spherical  A  RST  with  Z  R  =  Z  T. 
To  prove    r  =  t. 

ARGUMENT 

1.  Let  A  R's'T'  be  the  polar  of  A  SST. 

2.  Z  R  =  Z  r. 

3.  .-.  r'  =  t'. 

4.  .-.R^T1. 

5.  .-.    r  =  t.  Q.E.D. 


REASONS 

1.  §  943. 

2.  By  hyp. 

3.  §947. 

4.  §  961. 

5.  §  947. 


442 


SOLID   GEOMETRY 
PROPOSITION  XX.     THEOREM 


963.  //  two  angles  of  a  spherical  triangle  are  unequal, 
the  side  opposite  the  greater  angle  is  greater  than  the 
side  opposite  the  less  angle. 


Given    spherical  A  ABC  with  Z  A  >  Z  C. 
To  prove    BC>  AB. 

ARGUMENT 

1.  Draw  an  arc  of  a  great  O  AD  making 

Z  1  =  Z  C. 

2.  Then  ^Z>  =  £<7. 

3.  But  BD  -f  2)  >  ifl. 

4.  .  •.  S  +  Z>C'  >  2e ;  i.e.  BC  >  AB.    Q.E.D. 


REASONS 

1.  §  908,  g. 

2.  §  962. 

3.  §  941. 

4.  §  309. 


Ex.  1533.  In  a  birectangular  spherical  triangle  the  side  included  b] 
the  two  right  angles  is  less  than,  equal  to,  or  greater  than,  either  of 
other  two  sides,  according  as  the  angle  opposite  is  less  than,  equal  to,  01 
greater  than  90°. 

Ex.  1534.     An  equilateral  spherical  triangle  is  also  equiangular. 

Ex.  1535.     If  two  face  angles  of  a  trihedral  angle  are  equal,  the 
dihedral  angles  opposite  are  equal. 

Ex.  1536.     State  and  prove  the  converse  of  Ex.  1534. 

Ex.  1537.     State  and  prove  the  converse  of  Ex.  1535. 

Ex.  1538.     The  arcs  bisecting  the  base  angles  of  an  isosceles  spht 
cal  triangle  form  an  isosceles  spherical  triangle. 

Ex.  1539.  The  bases  of  an  isosceles  trapezoid  are  14  inches  and 
inches  and  the  altitude  3  inches ;  find  the  total  area  and  volume  of  tl 
solid  generated  by  revolving  the  trapezoid  about  its  longer  base  as  an  axis. 

Ex.  1540.     Find  the  total  area  and  volume  of  the  solid  generated  by 
revolving  the  trapezoid  of  Ex.  1539  about  its  shorter  base  as  an  axis. 


BOOK  IX  443 

PROPOSITION  XXI.     THEOREM 

964.   If  two  sides  of  a  spherical  triangle  are  unequal, 
the  angle  opposite  the  greater  side  is  greater  than  the 
angle  opposite  the  less  side. 
B 


Given   spherical  A  ABC  with  BC  >  AB. 

To  prove    Z  A  >  Z  (7. 

The  proof  is  left  as  an  exercise  for  the  student. 

HINT.     Prove  by  the  indirect  method. 

965.  Questions.  Could  Prop.  XXI  have  been  proved  by  the  method 
used  in  §  156  ?  Does  reason  4  of  that  proof  hold  in  spherical  A  ?  See 
Ex.  1525.  

Ex.  1541.  If  two  adjacent  sides  of  a  spherical  quadrilateral  are 
greater,  respectively,  than  the  other  two  sides,  the  spherical  angle  included 
between  the  two  shorter  sides  is  greater  than  the  spherical  angle  between 
the  two  greater  sides. 

HINT.     Compare  with  Ex.  153. 

Ex.  1542.  Find  the  total  area  and  volume  of  the  solid  generated  by 
revolving  the  trapezoid  of  Ex.  1539  about  the  perpendicular  bisector  of  its 
bases  as  an  axis. 

Ex.  1543.  Find  the  radius  of  the  sphere  in- 
scribed in  a  regular  tetrahedron  whose  edge  is  a. 

HINT.  Let  0  be  the  center  of  the  sphere,  A  the 
center  of  face  VED,  and  B  the  center  of  face  EDF. 
Then  OA  =  radius  of  inscribed  sphere.  Show  that 
rt.  A  VAO  and  VBC  are  similar.  Then  VO  :  VC 
=  VA  :  VS.  VC,  VA,  and  VB  can  be  found  (Ex. 
1328).  Find  FO,  then  OA. 

Ex.  1544.  Find  the  radius  of  the  sphere  circumscribed  about  a  regu- 
lar tetrahedron  whose  edge  is  a. 

HINT.   In  the  figure  of  Ex.  1543,  draw  AD  and  OD. 


444  SOLID   GEOMETRY 

MENSURATION   OF   THE   SPHERE 

AREAS 
PROPOSITION  XXII.     THEOREM 

966.  If  an  isosceles  triangle  is  revolved  about  a  straight 
line  lying  in  its  plane  and  passing  through  its  vertex  but 
not  intersecting  its  surface,  the  area  of  the  surf  ace  gener- 
ated by  the  base  of  the  triangle  is  equal  to  the  product 
of  its  projection  on  the  axis  and  the  circumference  of  a 
circle  whose  radius  is  the  altitude  of  the  triangle- 

X  X 


Given  isosceles  A  AOB  with  base  AB  and  altitude  OE,  a  str. 
line  XY  lying  in  the  plane  of  A  AOB  passing  through  0  and 
not  intersecting  the  surface  of  A  AOB,  and  CD  the  projection 
of  AB  on  XY;  let  the  area  of  the  surface  generated  by  AB  be 
denoted  by  area  AB. 

To  prove    area  AB  =  CD  •  2  irOE. 

I.   If  AB  is  not  II  XY  and  does  not  meet  XY  (Fig.  1). 

ARGUMENT  ONLY 

1.  From  E  draw  EH _L  XY. 

2.  Since  the  surface  generated  by  AB  is  the  surface  of  a 
frustum  of  a  rt.  circular  cone,  area  AB  =  AB  •  2  TrEH. 

3.  From  A  draw  AKA.  BD. 

4.  Then  in  rt.  A  BAK  and  OEH,  Z  BAK  —  /.  OEH. 


BOOK  IX 


445 


5.  .-.  ABAK~  AOEH. 

6.  .-.  AB  :  AK  =  OE :  EH',  i.e.  AB  -  EH  =  AK  -  OE. 

7.  But  AK  =  CD  ;   .'.  AB  -  EH=  CD  •  OE. 

8.  .*.  area  AB  =  CD  •  2  irOE.  Q.E.D. 

II.  If  AB  is  not  II  XY  and  point  A  lies  in  XY  (Fig.  2). 

III.  If  4511  JKT  (Fig.  3). 

The  proofs  of  II  and  III  are  left  as  exercises  for  the  student. 

HINT.     See  if  the  proof  given  for  I  will  apply  to  Figs.  2  and  3. 

967.  Cor.  I.  If  half  of  a  regular  polygon  with  an  even 
number  of  sides  is  circumscribed  about  a  semicircle,  the 
area  of  the  surface  generated  by  its  semiperimeter  as  it 
revolves  about  the  diameter  of 
the  semicircle  as  an  axis,  is  equal 
to  the  product  of  the  diameter 
of  the  regular  polygon  and  the 
circumference  of  a  circle  whose 
radius  is  E,  the  radius  of  the 
given  semicircle. 

OUTLINE  OF  PROOF 
1.   Area  4 '5'  =  A'F'  •  2  -n-R; 
area  B'c'  =  F'o'  -  2  irR;  etc. 

'B'C'-"  =  (A'F'  +  F'O'  + 


968.  Cor.  H.  If  half  of  a  regular  polygon  with  an  even 
number  of  sides  is  inscribed  in  a  semicircle,  the  area  of 
the  surface  generated  by  its  semi-  A 

perimeter  as  it  revolves  about 
the  diameter  of  the  semicircle  as 
an  axis,  is  equal  to  the  product  of 
the  diameter  of  the  semicircle 
and  the  circumference  of  a  circle 
whose  radius  is  the  apothem  of 
the  regular  polygon.  j) 

HINT.  Prove  area  AB (7  •••  =AE-2ira. 


446 


SOLID   GEOMETRY 


Cor.  m.  If  halves  of  regular  polygons  with  the 
same  even  number  of  sides  are  circumscribed  about,  and 
inscribed  in,  a  semicircle,  then  by  repeatedly  doubling  the 
number  of  sides  of  these  polygons,  and  making  the  poly- 
gons always  regular,  the  surfaces  generated  by  the  semi- 
perimeters  of  the  polygons  as  they  revolve  about  the 
diameter  of  the  semicircle  as  an  axis  approach  a  com- 
mon limit. 

OUTLINE  OF  PROOF 

1.  If  S  and  s  denote  the  areas  of 
the  surfaces  generated  by  the  semi- 
perimeters  A'B'c'-"  and  ABC---  as 
they  revolve  about  A'E'  as  an  axis, 
then  S  =  A'E'  -  2  -n-R  (§  967)  ; 
and  s  =  AE  •  2  na  (§  968). 

S  A'E'  •  2  -rrR  A'E'  R 
AE  a 
•  •  ~  polygon  ABC 


2.  .-.  5  = 

S          AE  -  2  -rra 

3.  But  polygon  A'B'C'> 


(§  438). 


4. 


5. 


6. 


A'E'       A'B' 


R 


...  ^L  =  ±±_  =  -  (§§  419,  435). 

AE         AB         a^ 


S-S 


R     R_ 

a     a 
R2-a2 


R2 


S  R2 

7.  .-.  by  steps  similar  to  Args.  6-11  (§  855),  S  and  s  ap- 
proach a  common  limit.  Q.E.D. 

970.  Def.  The  surface  of  a  sphere  is  the  common  limit 
which  the  successive  surfaces  generated  by  halves  of  regular 
polygons  with  the  same  even  number  of  sides  approach,  if  these 
eemipolygons  fulfill  the  following  conditions  : 

(1)  They  must  be  circumscribed  about,  and  inscribed  in,  a 
great  semicircle  of  the  sphere ; 

(2)  The  number  of   sides  must   be   successively   increased, 
each  side  approaching  zero  as  a  limit. 


BOOK  IX 
PROPOSITION  XXIII.     THEOREM 


447 


971.   The  area  of  the  surface  of  a  sphere  is  equal  to  four 
times  the  area  of  a  great  circle  of  the  sphere. 


Given   sphere  0  with  its  radius  denoted  by  R,  and  the  area 
of  its  surface  denoted  by  S. 
To  prove    S  =  4  TrJf?2. 


ARGUMENT 


1. 


2. 


In  the  semicircle  ACE  inscribe  ABODE, 
half  of  a  regular  polygon  with  an 
even  number  of  sides.  Denote  its 
apothem  by  a,  and  the  area  of  the 
surface  generated  by  the  semiperime- 
ter  as  it  revolves  about  AE  as  an  axis 
by  s'. 

Then  s'  =  AE  •  2  wa ; 

i.e.  S'  =  2  R  •  2  no,  =  4  irRa. 

3.  As  the  number  of  sides  of  the  regular 

polygon,  of  which  ABCDE  is  half,  is 
repeatedly  doubled,  S1  approaches  S 
as  a  limit. 

4.  Also  a  approaches  R  as  a  limit. 

.-.  4  TTR  -  a  approaches  4  irR  •  R,  i.e.  4  irR*, 

as  a  limit. 

But  S'  is  always  equal  to  4  nR  •  a. 
.:  S  =  4  TrR2.  Q.E.D. 


5. 

6. 

7. 


REASONS 
1.    §  517,  a. 


2.  §  968. 

3.  §970. 


4.  §  543,  I. 

5.  §  590. 

6.  Arg.  2. 

7.  §  355. 


448  SOLID   GEOMETRY 

972.  Cor.  I.  The  areas  of  the  surfaces  of  two  spheres  are 
to  each  other  as  the  squares  of  their  radii  and  as  the 
squares  of  their  diameters. 

OUTLINE  OF  PROOF 


S'         4  TTtf'2         R'* 

2    But  ^  =4*2  =  (2*)2  =D*          L  =  **- 
R'2      4*'2       (2R')2      D'2'      '  S'      D'*' 

973.  Historical  Note.  Prop.  XXIII  is  given  as  Prop.  XXXV  in 
the  treatise  entitled  Sphere  and  Cylinder  by  Archimedes,  already  spoken 
of  in  §  809. 


Ex.  1545.     Find  the  surface  of  a  sphere  whose  diameter  is  16  inches. 

Ex.  1546.  What  will  it  cost  to  gild  the  surface  of  a  globe  whose  radius 
is  1^  decimeters,  at  an  average  cost  of  f  of  a  cent  per  square  centimeter  ? 

Ex.  1547.  The  area  of  a  section  of  a  sphere  made  by  a  plane  11 
inches  from  the  center  is  3600  V  square  inches.  Find  the  surface  of  the 
sphere. 

Ex.  1548.  Find  the  surface  of  a  sphere  circumscribed  about  a  cube 
whose  edge  is  12  inches. 

Ex.  1549.  The  radius  of  a  sphere  is  It.  Find  the  radius  of  a  sphere 
whose  surface  is  twice  the  surface  of  the  given  sphere  ;  one  half  ;  one  nth. 

Ex.  1550.  Find  the  surface  of  a  sphere  whose  diameter  is  2  _R,  and 
the  total  surface  of  a  right  circular  cylinder  whose  altitude  and  diameter 
are  each  equal  to  2  It. 

Ex.  1551.  From  the  results  of  Ex.  1550  state,  in  the  form  of  a 
theorem,  the  relation  of  the  surface  of  a  sphere  to  the  total  surface  of  the 
circumscribed  cylinder. 

Ex.  1552.  Show  that,  in  Ex.  1550,  the  surface  of  the  sphere  is 
exactly  equal  to  the  lateral  surface  of  the  cylinder. 


974.  Historical  Note.  The  discovery  of  the  remarkable  property 
that  the  surface  of  a  sphere  is  two  thirds  of  the  surface  of  the  circum- 
scribed cylinder  (Exs.  1550  and  1551)  is  due  again  to  Archimedes.  The 
disco  very  of  this  proposition,  and  the  discovery  of  the  corresponding 
proposition  for  volumes  (§  1001),  were  the  philosopher's  chief  pride,  and 
he  therefore  asked  that  a  figure  of  this  proposition  be  inscribed  on  his 
tomb.  His  wishes  were  carried  out  by  his  friend  Marcellus.  (For  a 
further  account  of  Archimedes,  read  also  §  542.) 


BOOK   IX 


449 


975.  Defs.     A  zone  is  a  closed  figure  on  the  surface  of  a 
sphere  whose   boundary  is   composed   of  the   circumferences 
of  two  circles  whose  planes  are  parallel. 

The  circumferences  forming  the 
boundary  of  a  zone  are  its  bases. 

Thus,  if  semicircle  NES  is  revolved 
about  NS  as  an  axis,  arc  AB  will  gen- 
erate a  zone,  while  points  A  and  B 
will  generate  the  bases  of  the  zone. 

976.  Def.    The  altitude  of  a  zone 
is  the  perpendicular  from  any  point 
in  the  plane  of  one  base  to  the  plane 
of  the  other  base. 

977.  Def.   If  the  plane  of  one  of  the  bases  of  a  zone  is  tan- 
gent to  the  sphere,  the  zone  is  called  a  zone  of  one  base. 

Thus,  arc  NA  or  arc  RS  will  generate  a  zone  of  one  base. 

978.  Questions.    Is  the  term  "  zone  "  used  in  exactly  the  same  sense 
here  as  it  is  in  the  geography  ?     Name  the  geographical  zones  of  one  base  ; 
of  two  bases.      Name  the  five  circles  whose   circumferences   form   the 
bases  of  the  six  geographical  zones.     Which  of  these  are  great  circles  ?  * 

979.  Cor.  n.     The  area  of  a  zone  is  equal  to  the  product 
of  its  altitude  and  the  circumference  of  a  great  circle. 

OUTLINE  *OF  PROOF 

Let  S  denote  the  area  generated  by 
broken  line  A'B'C',  s  by  broken  line 
ABC,  and  Z  by  arc  ABC;  let  DE,  the 
altitude  of  the  zone,  be  denoted  by  H. 

Then      S  =  D'E' -  2 -n-R; 

S  =  DE  •  2  TTO,. 

.-.  -  =  ^r-      (See   Args.    2-5, 
s      a2 

§  969.)      Then   by   steps   similar    to 
§§  969-971,  Z  =  H-2TrR. 

*  The  student  will  observe  that  the  projection  used  in  this  figure  is  different  from  that 
used  in  the  other  figures. 


450  SOLID  GEOMETRY 

980.  Cor.  III.     In  equal  spheres,  or  in  the  same  sphere, 
the  areas  of  two  zones  are  to  each  other  as  their  altitudes. 

981.  Question.     In  general,  surfaces  are  to  each  other  as  the  prod- 
ucts of  two  lines.     Is  §  980  an  exception  to  this  rule  ?    Explain. 

Ex.  1553.  The  area  of  a  zone  of  one  base  is  equal  to  the  area  of  a 
circle  whose  radius  is  the  chord  of  the  arc  generating  the  zone. 

HINT.     Use  §§  979  and  444,  II. 

Ex.  1554.     Show  that  the  formula  of  §  971  is  a  special  case  of  §  979. 

Ex.  1555.  Find  the  area  of  the  surface  of  a  zone  if  the  distance 
between  its  bases  is  8  inches  and  the  radius  of  the  sphere  is  6  inches. 

Ex.  1556.  The  diameter  of  a  sphere  is  16  inches.  Three  parallel 
planes  divide  this  diameter  into  four  equal  parts.  Find  the  area  of  each 
of  the  four  zones  thus  formed. 

Ex.  1557.  Prove  that  one  half  of  the  earth's  surface  lies  within  30° 
of  the  equator. 

Ex.   1558.     Considering  the  earth  as  a  sphere  with  radius  .R,  find  the 

7?      27? 
area  of  the  zone  adjoining  the  north  pole,  whose  altitude  is  —  ;    -—  •     Is 

the  one  area  twice  the  other  ? 

Ex.  1559.  Considering  the  earth  as  a  sphere  with  radius  .R,  find  the 
area  of  the  zone  extending  30°  from  the  north  pole  ;  60°  from  the  north 
pole.  Is  the  one  area  twice  the  other  ? 

Ex.  1560.  Considering  the  earth  as  a  sphere  with  radius  J?,  find  the 
area  of  the  zone  whose  bases  are  parallels  of  latitude:  (a)  30°  and  45C 
from  the  north  pole  ;  (6)  30°  and  45°  from  the  equator.  Are  the  two 
areas  equal  ?  Explain  your  answer. 

Ex.  1561.  How  far  from  the  center  of  a  sphere  whose  radius  is  E 
must  the  eye  of  an  observer  be  so  that  one  sixth  of  the  surface  of  the 
sphere  is  visible  ? 

HINT.       Let  E  be    the  eye    of   the    observer. 

Then  AB  must  =^.     Find  OB,  then  use  §  443,  II. 
3 

Ex.  1562.  What  portion  of  the  surface  of  a 
sphere  can  be  seen  if  the  distance  of  the  eye  of  the 
observer  from  the  center  of  the  sphere  is  2E  ? 
37??  n  2t? 

Ex.  1563.  The  radii  of  two  concentric  spheres  are  6  inches  and  10 
inches.  A  plane  is  passed  tangent  to  the  inner  sphere.  Find :  (a)  the 
area  of  the  section  of  the  outer  sphere  made  by  the  plane  ;  (6)  the  area 
of  the  surface  cut  off  of  the  outer  sphere  by  the  plane. 


BOOK  IX 


451 


982.  Def.     A  lune  is  a  closed  figure  on  the  surface  of  a  sphere 
whose  boundary  is  composed   of  two 
semicircumferences  of  great  circles,  as 

NA8B. 

983.  Defs.     The  two  semicircumfer- 
ences are  called  the  sides  of  the  lune, 
as  NAS  and  NBS-,  the  points  of  inter- 
section  of    the    sides   are   called    the 
vertices  of  the  lune,  as  -ZV  and  S;  the 
spherical  angles  formed  at  the  vertices 
by  the  sides  of  the  lune  are  called  the 
angles  of  the  lune,  as  A  ANB  and  BSA. 

'984.   Prove,  by  superposition,  the  following  property  of  lunes : 
In  equal  spheres,  or  in  the  same  sphere,  two  lunes  are  equal  if 
their  angles  are  equal. 

985.  So  far,  the  surfaces  considered  in  connection  with  the 
sphere  have  been  measured  in  terms  of  square  units,  i.e.  square 
inches,  square  feet,  etc.    For  example,  if  the  radius  of  a  sphere 
is  6  inches,  the  surface  of  the  sphere  is  47r62,  i.e.  144  IT  square 
inches.     But,  as  the  sides  and  angles  of  a  lune  and  a  spherical 
polygon  are  given  in  degrees  and  not  in  linear  units,  it  will  be 
necessary  to  introduce  some  new  unit  for  determining  the  areas 
of  these  figures.    For  this  purpose  the  entire  surface  of  a  sphere 
is  thought  of  as  being  divided  into  720  equal  parts,  and  each 
one  of  these  parts  is  called  a  spherical  degree.     Hence : 

986.  Def.    A  spherical  degree  is  -^  of  the  surface  of  a  sphere. 

Now  if  the  area  of  a  lune  or  of  a  spherical  triangle  can  be  ob- 
tained in  spherical  degrees,  the  area  can  easily  be  changed  to 
square  units.  For  example,  if  it  is  found  that  the  area  of  a 
spherical  triangle  is  80  spherical  degrees,  its  area  is  -f^,  i.e.  ^ 
of  the  entire  surface  of  the  sphere.  On  the  sphere  whose 
radius  is  6  inches,  the  area  of  the  given  triangle  will  be  J-  of 
144  TT  square  inches,  i.e.  16  IT  square  inches.  The  following 
theorems  are  for  the  purpose  of  determining  the  areas  of  figures 
on  the  surface  of  a  sphere  in  terms  of  spherical  degrees. 


452 


SOLID   GEOMETRY 


PROPOSITION  XXIV.     THEOREM 

987.  The  area  of  a  lune  is  to  the  area  of  the  surface  of 
the  sphere  as  the  number  of  degrees  in  the  angle  of  the 
lune  is  to  360. 

N 


Q  E 


FIG.  1. 

Given  lune  NASB  with  the  number  of  degrees  in  its  Z  denoted 
by  N,  its  area  denoted  by  L,  and  the  area  of  the  surface  of  the 
sphere  denoted  by  S ;  let  O  EQ  be  the  great  O  whose  pole  is  N. 

L         N 

To  prove    -  =  — -• 
s      360 

I.  If  arc  AB  and  circumference  EQ  are  commensurable 
(Fig-  1). 

ARGUMENT  REASONS 

1.   Let  m  be  a  common  measure  of  arc  AB      1.    §  335. 
and  circumference  EQ,  and  suppose 
that  m  is  contained  in  arc  AB  r  times 
and  in  circumference  EQ  t  times, 
arc  AB  _  r 

t 


2.   Then 


circumference  EQ 

3.  Through  the  several  points  of  division 

on  circumference  EQ  pass  semicir- 
cumferences  of  great  circles  from  N 
to  S. 

4.  Then  lime  NASB  is  divided  into  r  lunes 

and  the  surface  of  the  sphere  into  t 
lunes,  each  equal  to  lune  NCSS, 


§341. 
§  908,  h. 


4.    §984. 


BOOK  IX 


453 


5. 


6. 


L_  r 

s  ~  t 

L 

S 


ARGUMENT 


arc  AB 


9. 


circumference  EQ 
But  arc  AB  is  the  measure  of 

it  contains  N  degrees. 
And  circumference  EQ  contains  360°. 
L         N 


i.e. 


Q.E.D. 


REASONS 

5.  §341. 

6.  §54,1. 

7.  §918. 

8.  §297. 

9.  §309. 


II.  If  arc  AB  and  circumference  EQ  are  incommensurable 
(Fig.  2). 

The  proof  is  left  as  an  exercise  for  the  student. 
HINT.     The  proof  is  similar  to  that  of  §  409,  II. 

988.  Cor.  I.  The  area  of  a  lune,  expressed  in  spherical 
degrees,  is  equal  to  twice  the  number  of  degrees  in  its 
angle. 

OUTLINE  OF  PROOF 


720       360 


Ex.  1564.  Find  the  area  of  a  lune  in  spherical  degrees  if  its  angle 
is  35°.  What  part  is  the  lune  of  the  entire  surface  of  the  sphere  ? 

Ex.  1565.  Find  the  area  of  a  lune  in  square  inches  if  its  angle  is  42° 
and  the  radius  of  the  sphere  is  8  inches.  (Use  ir  =  ^.) 

Ex.  1566.  In  equal  spheres,  or  in  the  same  sphere,  two  lunes  are  to 
each  other  as  their  angles. 

Ex.  1567.  Two  lunes  in  unequal  spheres,  but  with  equal  angles,  are  to 
each  other  as  the  squares  of  the  radii  of  their  spheres. 

Ex.  1568.  In  a  sphere  whose  radius  is  J?,  find  the  altitude  of  a  zone 
equivalent  to  a  lune  whose  angle  is  45°. 

Ex.  1569.  Considering  the  earth  as  a  sphere  with  radius  E,  find  the 
area  of  the  zone  visible  from  a  point  at  a  height  h  above  the  surface  of 
the  earth. 

989.  Def.  The  spherical  excess  of  a  spherical  triangle  is 
the  excess  of  the  sum  of  its  angles  over  180°. 


454 


SOLID   GEOMETRY 


PROPOSITION  XXV.     THEOREM 

990.    The  area  of  a   spherical   triangle,  expressed   in 
spherical  degrees,  is  equal  to  its  spherical  excess. 


Given   spherical  A  ABC  with,  its  spherical   excess   denoted 
by  E. 

To  prove    area  of  A  ABC  =  E  spherical  degrees. 


ARGUMENT 

1.  Complete  the  circumferences  of  which 

AB,  BC,  and  CA  are  arcs. 

2.  A  AB'c'  and  A'BC  are  symmetrical. 

3.  .'.  A  AB'C'  =0=  A  A'BC. 

4.  .-.  A  ABC  +  A  AB'C' o  A  ABC  +  A  A'BC. 

5.  .-.,  expressed  in  spherical  degrees, 

A  ABC  -f  A  AB'C'  =0  lune  A  =  2  A  ; 
A  ABC  +  A  AB'C  =  lune  B  =  2  B ; 
A  ABC  +  A  ABC'  =  lune  C  =  2  C. 
f>.    .-.  2AABC+(AABC+AAB'C'  +  AAB'C 
+  A  ABC')=  2(A  +  B+C). 

7.  EutAABC+AAB'C'+AAB'C+AABC' 

=  surface  of  a  hemisphere  =  360. 

8.  .-.  2  A  ABC-\-  360  =  2(^4-^  +  C). 

9.  .-.  A^J5(7+180=  4  +  #+  (7. 

10.    .-.A  ABC  =  (A  +  B  +  C)—  180,    i.e.     E 
spherical  degrees.  Q.E.D. 


REASONS 

1. 

§  908,  g. 

2. 

§  95G. 

3. 

§  958. 

4. 

§  54,  2. 

5. 

§  988. 

6. 

§  54,  2. 

7. 

§  985. 

8. 

§  309. 

9. 

§  54,  8  a. 

10. 

§  54,  3. 

BOOK  IX  455 

991.  In  §  949  it  was  proved  that  the  sum  of  the  angles  of 
a  spherical  triangle  is  greater  than  180°  and  less  than  540°. 
Hence  the  spherical  excess  of  a  spherical  triangle  may  vary 
from  0°  to  360°,  from  which  it  follows  (§  990)  that  the  area  of 
a  spherical  triangle  may  vary  from  y|^  to  ff-g-  of  the  entire 
surface  ;  i.e.  the  area  of  a  spherical  triangle  may  vary  from 
nothing  to  ^  the  surface  of  the  sphere.  Thus  in  a  spherical 
triangle  whose  angles  are  70°,  80°,  and  100°,  respectively,  the 
spherical  excess  is  (70°  +  80°  +  100°)  -  180°  =  70°  ;  i.e.  the 
area  of  the  given,  triangle  is  -f^  of  the  surface  of  the  sphere. 


992.  Historical  Note.  Menelaus  of  Alexandria  (circ.  98  A.D.) 
wrote  a  treatise  in  which  he  describes  the  properties  of  spherical  triangles, 
although  there  is  no  attempt  at  their  solution.  The  expression  for  the 
area  of  a  spherical  triangle,  as  stated  in  §  990,  was  first  given  about  1626 
A.D.  by  Girard.  (See  also  §  946.)  This  theorem  was  also  discovered  in- 
dependently by  Cavalieri,  a  prominent  Italian  mathematician. 


Ex.  1570.  If  three  great  circles  are  drawn,  each  perpendicular  to  the 
other  two,  into  how  many  trirectangular  spherical  triangles  is  the  surface 
divided  ?  Then  what  is  the  area  of  a  trirectangular  spherical  triangle  in 
spherical  degrees  ?  Test  your  answer  by  applying  Prop.  XXV. 

Ex.  1571.  Find  the  area  in  spherical  degrees  of  a  birectangular 
spherical  triangle  one  of  whose  angles  is  70°  ;  of  an  equilateral  spherical 
triangle  one  of  whose  angles  is  80°.  What  part  of  the  surface  of  the 
sphere  is  each  triangle  ? 

Ex.  1572.  The  angles  of  a  spherical  triangle  in  a  sphere  whose  sur- 
face has  an  area  of  216  square  feet  are  95°,  105°,  and  130°.  Find  the 
number  of  square  feet  in  the  area  of  the  triangle. 

Ex.  1573.  In  a  sphere  whose  diameter  is  16  inches,  find  the  area  of 
a  triangle  whose  angles  are  70°,  86°,  and  120°. 

Ex.  1574.  The  angles  of  a  spherical  triangle  are  60°,  120°,  and  160°, 
and  its  area  is  lOOf  square  inches.  Find  the  radius  of  the  sphere.  (Use 

»=¥•) 

Ex.  1575.  The  area  of  a  spherical  triangle  is  90  spherical  degrees, 
and  the  angles  are  in  the  ratio  of  2,  3,  and  5.  Find  the  angles. 

Ex.  1576.  Find  the  angle  (1)  of  an  equilateral  spherical  triangle, 
(2)  of  a  lune,  each  equivalent  to  one  third  the  surface  of  a  sphere. 

Ex.  1577.  Find  the  angle  of  a  lune  equivalent  to  an  equilateral 
spherical  triangle  one  of  whose  angles  is  84°. 


456 


SOLID   GEOMETRY 


PROPOSITION  XXVI.     THEOREM 

993.  The  area  of  a  spherical  polygon,  expressed  in 
spherical  degrees,  is  equal  to  the  sum  of  its  angles  dimin- 
ished by  180°  taken  as  many  times  less  two  as  the  poly- 
gon has  sides. 


Given  spherical  polygon  ABCD  •••  with  n  sides;  denote  the 
sum  of  its  angles  by  T. 

To  prove  area  of  polygon  ABCD  •  ••,  expressed  in  spherical 
degrees,  =  T—(n  —  2)180. 


ARGUMENT 

1.  From  any  vertex  such  as  A,  draw  all 

possible    diagonals    of   the    polygon, 
forming  n  —  2  spherical  A,  I,  II,  etc. 

2.  Then,  expressed  in  spherical  degrees, 

A  I  =  (Z  1  +  Z  2  +  Z  3)  -  180 ; 
AII=(Z  4-f-Z  5  +  Z  6) -180;  etc. 

3.  .-.  AI  +  AII+ ...  =  r-(n-2)180. 

4.  .-.    area    of    polygon    ABCD  •  •• 

=  r-(w  — 2)  180.  Q.E.D. 


1. 


REASONS 
§  937. 


2.   §  990. 


§  54,  2. 
§  309. 


Ex.  1578.  Prove  Prop.  XXVI  by  using  a  figure  similar  to  that  used 
in  §  216. 

Ex.  1579.  Find  the  area  of  a  spherical  polygon  whose  angles  are  80°, 
92°,  120°,  and  140°,  in  a  sphere  whose  radius  is  8  inches. 

Ex.  1580.  Find  the  angle  of  an  equilateral  spherical  triangle  equiva- 
lent to  a  spherical  pentagon  whose  angles  are  00°,  100°,  110°,  130°,  and  140°. 

Ex.  1581.  Find  one  angle  of  an  equiangular  spherical  hexagon 
equivalent  to  six  equilateral  spherical  triangles  each  with  angles  of  70°. 


BOOK   IX 


457 


1 


K---.-X 


1 


Ex.  1582.  The  area  of  a  section  of  a  sphere  63  inches  from  the  cen- 
ter is  256  TT  square  inches.  Find  the  surface  of  the  sphere. 

Ex.  1583.  The  figure  represents  a  sphere 
inscribed  in  a  cylinder,  and  two  cones  with 
the  bases  of  the  cylinder  as  their  bases  and  the 
center  of  the  sphere  as  their  vertices.  Any 
plane,  as  BS,  is  passed  through  the  figure 
parallel  to  MN,  the  plane  of  the  base.  Prove 
that  the  ring  between  section  AD  of  the 
cylinder,  and  section  CD  of  the  cone,  is  always 
equivalent  to  the  section  of  the  sphere. 

Ex.  1584.  Find  the  volume  of  a  barrel  30  inches  high,  54  inches  in 
circumference  at  the  top  and  bottom,  and  64  inches  in  circumference  at 
the  middle. 

HIXT.     Consider  the  barrel  as  the  sum  of  two  frustums  of  cones. 

Ex.  1585.  Given  T  the  total  area,  and  E  the  radius  of  the  base,  of 
a  right  circular  cylinder.  Find  the  altitude. 

Ex.  1586.  Given  $the  lateral  area,  and  E  the  radius  of  the  base,  of 
a  right  circular  cone.  Find  the  volume. 

Ex.  1587.  Given  8  the  lateral  area,  and  T  the  total  area,  of  a  right 
circular  cone.  Find  the  radius  and  the  altitude. 


VOLUMES 

994.  Note.  The  student  should  not  fail  to  observe  the  striking 
similarity  in  the  figures  and  theorems,  as  well  as  in  the  definitions,  relat- 
ing to  the  areas  and  volumes  connected  with  the  measurement  of  the 
sphere.  A  careful  comparison  of  the  following  articles  will  emphasize 
this  similarity : 


AREAS 

§966 

>§  967,  968 
§969 
§970 
§971 
§972 
§975 
§977 


VOLUMES 

§995 
§996,  a 
§  996,  6 
§  996,  c 
§997 
§999 
§  1002 
8  1003 


AREAS 

§979 
§982 
§984 
§987 
§988 
§  936 
§  990 


VOLUMES 
5§  1004,  1005 
§  1006 
§1007 
§  1008 
§  1009 
§  1010 
§1012 
§  1013 


458 


SOLID   GEOMETRY 


PROPOSITION  XXVII.     THEOREM 

995.  If  an  isosceles  triangle  is  revolved  about  a  straight 
line  lying  in  its  plane  and  passing  through  its  vertex  but 
not  intersecting  its  surface,  the  volume  of  the  solid  gener- 
ated is  equal  to  the  product  of  the  surface  generated  by 
the  base  of  the  triangle  and  one  third  of  its  altitude. 


—\-D 


E 


B 


0 


Y 

FIG.  2.  FIG.  3. 

Given  isosceles  A  A  OB  with  altitude  OE,  and  a  str.  line  XY 
lying  in  the  plane  of  A  AOB,  passing  through  0  and  not  inter- 
secting the  surface  of  A  AOB;  let  the  volume  of  the  solid  gen- 
erated by  A  AOB  revolving  about  XY  as  an  axis  be  denoted  by 
volume  AOB. 

To  prove   volume  AOB  =  area  AB  •  -J  OE. 

I.    If  AB  is  not  II  XY  and  does  not  meet  XY  (Fig.  1). 

ARGUMENT  ONLY 

1.  Draw  AC  and  BD  _L  XY. 

2.  Prolong  BA  to  meet  XY  at  F. 

3.  Then  volume  .40.B  =  volume  FOB  —  volume  FOA. 

4.  Volume  FOB  =  volume  FDB  +  volume  DOB. 

.    5.   .-.   volume  FOB  =  ^irlsi?  -  FD  +  I  TT £Z>2  •  -DO 

=  1  TT  BD\FD  +  £0)  =  ^irBD  -  BD 
But        BD  -  FO  —  twice  area  of  A  ^OJ5  =  .ay  •  0#. 
.-.  volume    FOB  =  %  TT  BD  •  BF  •  OE  =  TT  BD  -  BF  •  J  OE. 
But      7r##  •  BF  =  area  ^^. 
.-.  volume    FOB  =  area  FB  •  i  O.E. 


FO. 


6. 

7. 
8. 
9. 


BOOK   IX 


459 


10.  Likewise  volume  FOA  =  area  FA  •  1  OE. 

o 

11.  .-.  volume  AOB  =  area  FB  •  ±  OE  —  area  FA  -  ^  OE 

=  (area  FB  —  area  FA)  ±  OE 

=  area  AB  •  i  OE.  Q.E.D. 

II.  If  AB  is  not  II  XY  and  point  A  lies  in  XY  (Fig.  2). 
The  proof  is  left  as  an  exercise  for  the  student.* 
HINT.     See  Arg.  9  or  Arg.  10  of  §  995,  I. 

III.  If  AB  II  XY  (Fig.  3). 

The  proof  is  left  as  an  exercise  for  the  student. 
HINT.     Volume  AOB  =  volume  ACDB  —  twice  volume  CO  A. 

996.   The  student  may  : 

(a)  State  and  prove  the  corollaries  on  the  volume  of  a  sphere 
corresponding  to  §§  967  and  968. 

(&)  State  and  prove  the  theorem  on  the  volume  of  a  sphere 
corresponding  to  §  969. 


1. 


3. 


OUTLINE  OF  PROOF 

V—  (area  A'B'C'  •••^R 
=  A'E'  >27rR-±R 

—    2          J?2  A  f  T?rf 

— — '  ~o"  TTXt     •   ./I  Ji 

(§§  996,  a  and  967).   C'< 
v  =  (area  4.B  (7  •  •  •  )|-  a 


996,  a  and  968). 
*      A'E' 
2       AE 


B' 


D 


v       --  Tra  -  AE       a        AE       a3 

4.  Proceed  as  in  §  855,  observing  that  since  the  limit  of 
«=/?(§  543,  I),  the  limit  of  as=  R3  (§  593);  i.e.  R3-a3  may 
be  made  less  than  any  previously  assigned  value,  however 
small. 

(c)  State,  by  aid  of  §  970,  the  definition  of  the  volume  of  a 
sphere. 


460 


SOLID   GEOMETRY 


PROPOSITION  XXVIII.     THEOREM 

997.    The  volume  of  a  sphere  is  equal  to  the  product 
of  the  area  of  its  surface  and  one  third  its  radius. 


B 


Given   sphere  0  with  its  radius  denoted  by  R,  the  area  of  its 
surface  by  fif,  and  its  volume  by  F. 
To  prove    V=S-±R. 


ARGUMENT 

1.  In  the  semicircle  ACE  inscribe  ABODE, 

half  of  a  regular  polygon  with  an  even 
number  of  sides.  Denote  its  apothem 
by  a,  the  area  of  the  surface  generated 
by  the  semiperimeter  as  it  revolves 
about  AE  as  an  axis  by  s',  and  the  vol- 
ume of  the  solid  generated  by  semi- 
polygon  A  B  ODE  by  F'. 

2.  Then  V1  =  S'  •  J  a. 

3.  As  the  number  of  sides  of  the  regular 

polygon,  of  which  ABODE  is  half,  is 
repeatedly  doubled,  V1  approaches  F 
as  a  limit. 

4.  Also  s'  approaches  S  as  a  limit. 

5.  And  a  approaches  R  as  a  limit. 

6.  .'.  S'  •  a  approaches  S  •  R  as  a  limit. 

7.  .'.  S1  '  ^  a  approaches  S  •  ^  R  as  a  limit. 

8.  But  V  is  always  equal  to  s'  •  J«. 

9.  /.  V=S  •  I  R.      •  Q.K.I). 


REASONS 
§  517,  a. 


§  996,  a. 
§  996,  c. 


§  970. 
§  543,  I. 
§  592. 
§  590. 
Arg.  2. 


BOOK  IX  461 

998.  Cor.  I.     If  v  denotes  the  volume,  R  the  radius,  and 
D  the  diameter  of  a  sphere, 

V=±TTR*=lirD*. 

999.  Cor.  n.     The  volumes  of  two  spheres  are  to  each 
other  as  the  cubes  of  their  radii  and  as  the  cubes  of  their 
diameters.    (HINT.    See  §  972.) 

1000.  Historical  Note.     It  is  believed  that  the  theorem  of  §  999 
was  proved  as  early  as  the  middle  of  the  fourth  century  B.C.  by  Eudoxus, 
a  great  Athenian  mathematician  already  spoken  of  in  §§  809  and  896. 

Ex.  1588.  Find  the  volume  of  a  sphere  inscribed  in  a  cube  whose 
edge  is  8  inches. 

Ex.  1589.  The  volume  of  a  sphere  is  1774|  TT  cubic  centimeters.  Find 
its  surface. 

Ex.  1590.  Find  the  radius  of  a  sphere  equivalent  to  a  cone  with  alti- 
tude a  and  radius  of  base  6. 

Ex.  1591.  Find  the  radius  of  a  sphere  equivalent  to  a  cylinder  with 
the  same  dimensions  as  those  of  the  cone  in  Ex.  1590. 

Ex  1592.  The  metal  cone  and  cylinder  in  the  figure  have  their  alti- 
tude and  diameter  each  equal  to  2  B,  the  diameter  of  the  sphere.  Place 


the  sphere  in  the  cylinder,  then  fill  the  cone  with  water  and  empty  it  into 
the  cylinder.  How  nearly  is  the  cylinder  filled  ?  Next  fill  the  cone  with 
water  and  empty  it  into  the  cylinder  three  times.  Is  the  cylinder  filled  ? 
Ex.  1593.  From  the  results  of  Ex.  1592  state,  in  the  form  of  a 
theorem,  the  relation  of  the  volume  of  a  sphere :  (a)  to  the  volume  of  a 
circumscribed  cylinder;  (6)  to  the  volume  of  the  corresponding  cone. 
Prove  these  statements. 

1001.  Historical  Note.  The  problem  "To  find  a  sphere  equivalent 
to  a  given  cone  or  a  given  cylinder"  (Exs.  1590  and  1591),  as  well  as  the 
properties  that  the  volume  of  a  sphere  is  two  thirds  of  the  volume  of  the 
circumscribed  cylinder  and  twice  the  volume  of  the  corresponding  cone 
(Exs.  1592  and  1593),  are  due  to  Archimedes.  The  importance  attached 
to  this  by  the  author  himself  is  spoken  of  more  fully  in  §§  542  and  974. 


462 


SOLID   GEOMETRY 


Ex.  1594.  A  bowl  whose  inner  surface  is  an  exact  hemisphere  is 
made  to  hold  |  gallon  of  water.  Find  the  diameter  of  the  bowl. 

Ex.  1595.  A  sphere  12  inches  in  diameter  weighs,  93  pounds.  Find 
the  weight  of  a  sphere  of  the  same  material  16  inches  in  diameter. 

Ex.  1596.  In  a  certain  sphere  the  area  of  the  surface  and  the  volume 
have  the  same  numerical  value.  Find  the  volume  of  the  sphere. 

Ex.  1597.  Find  the  volume  of  a  spherical  shell  5  inches  thick  if  the 
radius  of  its  inner  surface  is  10  inches. 

Ex.  1598.  A  pine  sphere  24  inches  in  diameter  weighs  175  pounds. 
Find  the  diameter  of  a  sphere  of  the  same  material  weighing  50  pounds. 

Ex.  1599.  The  radius  of  a  sphere  is  H.  Find  the  radius  of  a  sphere 
whose  volume  is  one  half  the  volume  of  the  given  sphere ;  twice  the  vol- 
ume ;  n  times  the  volume. 

1002.   Defs.     A  spherical  sector  is  a  solid  closed  figure  gen- 
erated by  a  sector  of  a  circle  revolving  about  a  diameter  of  the 
circle  as  an  axis. 
C 


FIG.  2. 


FIG.  3. 


The  zone  generated  by  the  arc  of  the  circular  sector  is  called 
the  base  of  the  spherical  sector. 

1003.  Def.  If  one  radius  of  the  circular  sector  generating 
a  spherical  sector  is  a  part  of  the  axis,  i.e.  if  the  base  of  the 
spherical  sector  is  a  zone  of  one  base,  the  spherical  sector  is 
sometimes  called  a  spherical  cone. 

Thus  if  circular  sector  AOB  (Fig.  1)  revolves  about  diameter 
CD  as  an  axis,  arc  AB  will  generate  a  zone  which  will  be  the 
base  of  the  spherical  sector  generated  by  circular  sector  AOB 
(Fig.  2).  If  circular  sector  HOC  revolves  about  diameter  CDy 
the  spherical  sector  generated,  whose  base  is  the  zone  generated 
by  arc  BC,  will  be  a  spherical  cone  (Fig.  3). 


BOOK   IX 


463 


1004.  Cor.  HE.  The  volume  of  a  spherical  sector  is 
equal  to  the  product  of  its  base  and  one  third  the  radius 
of  the  sphere. 

OUTLINE  OF  PROOF 

Let  V  denote  the  volume  generated  by  polygon  OA'B'c1,  v  the 
volume  generated  by  polygon  OABC,  S  the  area  of  the  surface 
generated  by  broken  line  A'B'C',  s  the 
area    of   the    surface    generated    by 
broken  line  ABC,  and  Z  the  base  of 
the  spherical  sector  generated  by  cir- 
cular sector  AOC. 

Then  V  =  S  - 1  R,  and  v  =  s  •  4  a. 


But  -  =  —y  (Args.  2-5,  §  969). 

S         CL 


R?    R 

a2    a 


Then  by  steps  similar  to  §  996,  b  and  c,  and  §  997,  the 
volume  of  the  spherical  sector  generated  by  circular  sector 
AOC  =  Z  •  J  R. 

1005.  Cor.  IV.  If  v  denotes  the  volume  of  a  spherical 
sector,  z  the  area  of  the  zone  forming  its  base,  H  the  alti- 
tude of  the  zone,  and  R  the  radius  of  the  sphere, 

V=Z-\R  (§1004) 

(§979) 


Ex.  1600.     Considering  the  earth  as  a  sphere  with  radius  B,  find  the 
volume  of  the  spherical  sector  whose  base  is  a  zone  adjoining  the  north 


7?      97? 

pole  and  whose  altitude  is  —  ;  -™ 
3       3 


Is  the  one  volume  twice  the  other? 


Compare  your  results  with  those  of  Ex.  1558. 

Ex.  1601.  Considering  the  earth  as  a  sphere  with  radius  22,  find  the 
volume  of  the  spherical  sector  whose  base  is  a  zone  extending  :  (a)  30° 
from  the  north  pole  ;  (ft)  60°  from  the  north  pole.  Is  the  one  volume 
twice  the  other  ?  Compare  your  results  with  those  of  Ex.  1559. 


464  SOLID   GEOMETRY 

Ex.  1602.  Considering  the  earth  as  a  sphere  with  radius  J?,  find  the 
volume  of  the  spherical  sector  whose  base  is  a  zone  lying  between  the 
parallels  of  latitude :  (a)  30°  and  45°  from  the  north  pole ;  (6)  30°  and 
45°  from  the  equator.  Are  the  two  volumes  equal  ?  Compare  your 
results  with  those  of  Ex.  1560. 

Ex.  1603.  Considering  the  earth  as  a  sphere  with  radius  R,  find  the 
area  of  the  zone  whose  bases  are  the  circumferences  of  small  circles,  one 
30°  north  of  the  equator,  the  other  30°  south  of  the  equator.  What  part 
of  the  entire  surface  is  this  zone  ? 

Ex.  1604.  '  What  part  of  the  entire  volume  of  the  earth  is  that  por- 
tion included  between  the  planes  of  the  bases  of  the  zone  in  Ex.- 1603  ? 

HINT.     This  volume  consists  of  two  pyramids  and  a  spherical  sector. 

Ex.  1605.  A  spherical  shell  2  inches  in  thickness  contains  the  same 
amount  of  material  as  a  sphere  whose  radius  is  6  inches.  Find  the  radius 
of  the  outer  surface  of  the  shell. 

Ex.  1606.  A  spherical  shell  3  inches  thick  has  an  outer  diameter  of 
16  inches.  Find  the  volume  of  the  shell. 

Ex.  1607.  Find  the  volume  of  a  sphere  circumscribed  about  a  rec- 
tangular parallelepiped  whose  edges  are  3,  4,  and  12. 

Ex.  1608.  Find  the  volume  of  a  sphere  inscribed  in  a  cube  whose 
volume  is  686  cubic  centimeters. 

Ex.  1609.  The  surface  of  a  sphere  and  the  surface  of  a  cube  are  each 
equal  to  8.  Find  the  ratio  of  their  volumes.  Which  is  the  greater  ? 

Ex.  1610.  In  a  certain  sphere  the  volume  and  the  circumference  of  a 
great  circle  have  the  same  numerical  value.  Find  the  surface  and  the  vol- 
ume of  the  sphere. 

Ex.  1611.  How  many  bullets  \  of  an  inch  in  diameter  can  be  made 
from  a  sphere  of  lead  10  inches  in  diameter  ?  from  a  cube  of  lead  whose 
edge  is  10  inches  ? 

1006.  Defs.     A  spherical  -wedge  is  a  solid  closed  figure  whose 
bounding  surface  consists  of  a  June  and  the  planes  of  the  sides 
of  the  lune. 

The  lune  is  called  the  base  of  the  spherical  wedge,  and  the 
angle  of  the  lime  the  angle  of  the  spherical  wedge. 

1007.  Prove,  by  superposition,  the   following   property   of 
wedges : 

In  equal  spheres,  or  in  the  same  sphere,  two  spherical  wedges 
are  equal  if  their  angles  are  equal. 


BOOK   IX 


465 


PROPOSITION  XXIX.     THEOREM 

1008.  The  volume  of  a  spherical  wedge  is  to  the  volume 
of  the  sphere  as  the  number  of  degrees  in  the  angle  of 
the  spherical  wedge  is  to  360. 

N  N 


Q  E 


Given  spherical  wedge  NASB  with  the  number  of  degrees  in 
its  Z  denoted  by  N,  its  volume  denoted  by  TT,  and  the  volume 
of  the  sphere  denoted  by  F;  let  O  EQ  be  the  great  O  whose 
pole  is  N. 

W       N 
To  prove    —  = . 

v      360 

The  proof  is  left  as  an  exercise  for  the  student. 
HINT.     The  proof  is  similar  to  that  of  §  987. 

1009.  Cor.  I.  The  volume  of  a  spherical  wedge  is 
equal  to  the  product  of  its  base  and  one  third  the  radius 
of  the  sphere. 

OUTLINE  OF  PROOF 


S  •  i  R  =  L  •  i  R,  where  S  represents 


•  w  =  -*-.  r  =  -*- 
360  360 

the  area  of  the  surface  of  the  sphere,  and  L  the  area  of  the 
lune,  i.e.  the  area  of  the  base  of  the  spherical  wedge. 


Ex.  1612.     Tn  a  sphere  whose  radius  is  16  inches,  find  the  volume  of 
a  spherical  wedge  whose  angle  is  40°. 


466 


SOLID   GEOMETRY 


1010.  Defs.      A   spherical  pyramid    is   a   solid   closed  figure 
whose  bounding  surface  consists  of  a  spherical  polygon  and 
the  planes  of  the  sides  of  the  spherical 

polygon.      The    spherical    polygon   is          /^ 
the  base,  and  the  center  of  the  sphere 
the  vertex,  of  the  spherical  pyramid. 

1011.  By  comparison  with  §  957,  b 
and  §  958,  prove  the  following  prop- 
erty of  spherical  pyramids : 

In  equal  spheres,  or  in  the  same  sphere, 
two  triangular  spherical  pyramids  whose 
bases  are  symmetrical  spherical  triangles  are  equivalent. 

1012.  Cor.  II.     The  volume  of  a  spherical   triangular 
pyramid  is  equal  to  the  product  of  its  base  and  one  third 
the  radius  of  the  sphere. 

OUTLINE  OF  PROOF 

1.  Pyramid  0-AB'c'  o  pyramid 
O-A'BC  (§  1011). 

2.  .-.  pyramid  0-ABC  -f  pyramid    < 
0-AB'c'   =o   wedge    A  =  2  A  •  1  R 

(§  1009)  ;  pyramid  0-ABC  +  pyra- 
mid O-AB'C  =  wedge  B  =  2  B  -  ifl; 
pyramid  0—ABC  -f-  pyramid  0—ABC' 
=  wedge  C  =  2  c  -  ^  R. 

3.  .-.  twice  pyramid  0-ABC  +  hemispnere  =  2  (A  -f-  B  +  C)  1  R. 

4.  .-.  twice  pyramid  0-ABC  -+-  360  •  1  R  =  2(A  -f  B  -f  c) ±  R. 

5.  .-.  pyramid  0-ABC  =  (A  +  B  -f  C  —  180)  |  R 

=  A  ABC  •  ^  R  =  K  •  ±R.  Q.E.D. 

1013.  Cor.  in.     The  volume  of  any  spherical  pyramid 
is  eqiuyl  to  the  product  of  its  base  and  one  third  the  ra- 
dius of  the  sphere.     (HINT.     Compare  with  §805.) 

Ex.  1613.  Show  that  the  formula  of  §  007  is  a  special  case  of  §§  1004, 
1000  and  1013. 

Ex.  1614.  In  a  sphere  whose  radius  is  12  inches,  find  the  volume  of 
a  spherical  pyramid  whose  base  is  a  triangle  with  angles  70°,  80°,  and  00°. 


BOOK  IX 


467 


MISCELLANEOUS  EXERCISES  ON  SOLID  GEOMETRY 

Ex.  1615.  A  spherical  pyramid  whose  base  is  an  equiangular  penta- 
gon is  equivalent  to  a  wedge  whose  angle  is  30°.  Find  an  angle  of  the 
base  of  the  pyramid. 

Ex.  1616.  The  volume  of  a  spherical  pyramid  whose  base  is  an  equi- 
angular spherical  triangle  with  angles  of  105°  is  128  IT  cubic  inches.  Find 
the  radius  of  the  sphere. 

Ex.  1617.  In  a  sphere  whose  radius  is  10  inches,  find  the  angle  of  a 
spherical  wedge  equivalent  to  a  spherical  sector  whose  base  has  an  alti- 
tude of  12  inches. 

Ex.  1618.  Find  the  depth  of  a  cubical  tank  that  will  hold  100  gallons 
of  water. 

Ex.  1619.  The  altitude  of  a  pyramid  is  H.  At  what  distance  from 
the  vertex  must  a  plane  be  passed  parallel  to  the  base  so  that  the  part  cut 
off  is  one  half  of  the  whole  pyramid  ?  one  third  ?  one  nth  ? 

Ex.  1620.  Allowing  550  pounds  of  copper  to  a  cubic  foot,  find  the 
weight  of  a  copper  wire  \  of  an  inch  in  diameter  and  2  miles  long. 

Ex.  1621.  Disregarding  quality,  and  considering  oranges  as  spheres, 
i.e.  as  similar  solids,  determine  which  is  the  better  bargain,  oranges 
averaging  2f  inches  in  diameter  at  15  cents  per  dozen,  or  oranges  averaging 
3^  inches  in  diameter  at  30  cents  per  dozen. 

Ex.  1622.  In  the  figure,  J5,  O,  and  D  are 
the  mid-points  of  the  edges  of  the  cube  meet- 
ing at  A.  What  part  of  the  whole  cube  is  the 
pyramid  cut  off  by  plane  BCD  ? 

HINT.  Consider  ABC  as  the  base  and  D  as 
the  vertex  of  the  pyramid. 

Ex.  1623.  Is  the  result  of  Ex.  1622  the 
same  if  the  figure  is  a  rectangular  parallelepiped  ?  any  parallelepiped  ? 

Ex.  1624.  It  is  proved  in  calculus  that  in  order  that  a  cylindrical  tin 
can  closed  at  the  top  and  having  a  given  capacity  may  require  the  small- 
est possible  amount  of  tin  for  its  construction,  the  diameter  of  the  base 
must  equal  the  height  of  the  can.  Find  the  dimensions  of  such  a  can 
holding  1  quart ;  2  gallons. 

Ex.  1625.  A  cylindrical  tin  can  holding  2  gallons  has  its  height  equal 
to  the  diameter  of  its  base.  Another  cylindrical  tin  can  with  the  same 
capacity  has  its  height  equal  to  twice  the  diameter  of  its  base.  Find  the 
ratio  of  the  amount  of  tin  required  for  making  the  two  cans.  Is  your 
answer  consistent  with  the  fact  contained  in  Ex.  1624  ? 


468  SOLID   GEOMETRY 

Ex.  1626.  A  cannon  ball  12  inches  in  diameter  is  melted,  and  the  lead 
is  cast  in  the  form  of  a  cube.  Find  the  edge  of  the  cube. 

Ex.  1627.  The  cube  of  Ex.  1626  is  melted,  and  the  lead  is  cast  in  the 
form  of  a  cone,  the  diameter  of  whose  base  is  12  inches.  Find  the  altitude 
of  the  cone. 

Ex.  1628.  Find  the  weight  of  the  cannon  ball  in  Ex.  1626  if  a  cubic 
foot  of  iron  weighs  450  pounds. 

Ex.  1629.  The  planes  determined  by  the  diagonals  of  a  cube  divide 
the  cube  into  six  equal  pyramids. 

Ex.  1630.  Let  D,  E,  F,  and  G  be  the  mid-points  of  VA,  AB,  EC, 
and  OF,  respectively,  of  triangular  pyramid  V-ABC.  Prove  DEFG  a 
parallelogram. 

Ex.  1631.  In  the  figure,  is  plane  DEFG  par- 
allel to  edge  AC?  to  edge  VB?  Prove  that  any 
section  of  a  triangular  pyramid  made  by  a  plane 
parallel  to  two  opposite  edges  is  a  parallelogram. 

Ex.  1632.  The  three  lines  joining  the  mid- 
points of  the  opposite  edges  of  a  tetrahedron  bisect 
each  other  and  hence  meet  in  a  point. 

HINT.     Draw  DF  and  EG.     Are  these  two  of  the  required  lines  ? 

Ex.  1633.  In  a  White  Mountain  two-quart  ice  cream  freezer,  the 
can  is  4f-  inches  in  diameter  and  Q\  inches  high  ;  the  tub  is  6|  inches  in 
diameter  at  the  bottom,  8  inches  at  the  top,  and  9f  inches  high,  inside 
measurements,  (a)  Does  the  can  actually  hold  2  quarts  ?  (6)  How 
many  cubic  inches  of  ice  can  be  packed  about  the  can  ? 

Ex.  1634.  Find  the  total  area  of  a  regular  tetrahedron  whose  alti- 
tude is  a  centimeters. 

Ex.  1635.  The  lateral  faces  of  a  triangular  pyramid  are  equilateral 
triangles,  and  the  altitude  of  the  pyramid  is  6  inches.  Find  the  total  area. 

Ex.  1636.  In  the  foundation  work  of  the  Woolworth  Building,  a  55- 
story  building  on  Broadway,  New  York  City,  it  was  necessary,  in  order 
to  penetrate  the  sand  and  quicksand  to  bed  rock,  to  sink  the  caissons  that 
contain  the  huge  shafts  of  concrete  to  a  depth,  in  some  instances,  of  131 
feet.  If  the  largest  circular  caisson,  19  feet  in  diameter,  is  130  feet  deep 
and  was  filled  with  concrete  to  within  30  feet  of  the  surface,  how  many 
loads  of  concrete  were  required,  considering  1  cubic  yard  to  a  load  ? 

Ex.  1637.  From  A  draw  a  line  meeting  line  XY  in  B  ;  let  G  be  the 
mid-point  of  AB.  Find  the  locus  of  C  as  B  moves  in  line  XY. 

Ex.  1638.  In  Ex.  1637,  let  XY  be  a  plane.  Find  the  locus  of  C  as  B 
moves  arbitrarily  in  plane  XY. 


BOOK  IX  469 

Ex.  1639.  A  granite  shaft  in  the  form  of  a  frustum  of  a  square 
pyramid  contains  161|  cubic  feet  of  granite  ;  the  edges  of  the  bases  are  4 
feet  and  1£  feet,  respectively.  Find  the  height  of  the  shaft. 

Ex.  1640.  The  volume  of  a  regular  square  pyramid  is  42f  cubic  feet ; 
its  altitude  is  twice  one  side  of  the  base,  (a)  Find  the  total  surface  of 
the  pyramid ;  (6)  find  the  area  of  a  section  made  by  a  plane  parallel  to 
the  base  and  one  foot  from  the  base. 

Ex.  1641.     Allowing  1  cubic  yard  to  a  load,  find  the  number  of  loads 
of  earth  in  a  railway  cut  \  mile  in  length,  the  average  dimensions  of  a 
cross  section  being  as  represented  in  the  figure, 
the  numbers  denoting  feet.     Give  the  name  of 
the  geometrical  solid  represented  by  the  cut. 
Why  is  it  not  a  frustum  of  a  pyramid  ? 

Ex.  1642.  For  protection  against  fire,  a  tank  in  the  form  of  a  frustum 
of  a  right  circular  cone  was.  placed  in  the  tower  room  of  a  certain  public 
building.  The  tank  is  16  feet  in  diameter  at  the  bottom,  12  feet  in  di- 
ameter at  the  top,  and  16  feet  deep.  If  the  water  in  the  tank  is  never 
allowed  to  get  less  than  14  feet  deep,  how  many  cubic  feet  of  water  would 
be  available  in  case  of  an  emergency  ?  how  many  barrels,  counting  4^ 
cubic  feet  to  a  barrel  ? 

Ex.  1643.  A  sphere  with  radius  E  is  inscribed  in  a  cylinder,  and  the 
cylinder  is  inscribed  in  a  cube.  Find  :  (a)  the  ratio  of  the  volume  of 
the  sphere  to  that  of  the  cylinder ;  (&)  the  ratio  of  the  cylinder  to  the 
cube  ;  (c)  the  ratio  of  the  sphere  to  the  cube. 

Ex.  1644.  A  cone  has  the  same  base  and  altitude  as  the  cylinder  in 
Ex.  1643.  Find  the  ratio  of  the  cone  :  (a)  to  the  sphere  ;  (6)  to  the 
cylinder;  (c)  to  the  cube. 

Ex.  1645.  In  a  steam-heated  house  the  heat  for  a  room  was  supplied 
by  a  series  of  10  radiators  each  3  feet  high.  gl 

The  average  cross  section  of  a  radiator  is     s\ ~ "TN 

shown  in  the  figure,  the  numbers  denoting    yj1^ *\J 

inches.     It  consists  of  a  rectangle  with  a 

semicircle  at  each  end.    Find  the  total  radiating  surface  in  the  room. 

Ex.  1646.  A  coffee  pot  is  5  inches  deep,  4£  inches  in  diameter  at  the 
top,  and  5|  inches  in  diameter  at  the  bottom.  How  many  cups  of  coffee 
will  it  hold ,  allowing  6  cups  to  a  quart  ?  (Answer  to  nearest  whole  number. ) 

Ex.  1647.  Any  plane  passing  through  the  center  of  a  parallelepiped 
divides  it  into  two  equivalent  solids.  Are  these  solids  equal  ? 

Ex.  1648.  From  two  points,  P  and  _Z?,  on  the  same  side  of  plane  AB, 
two  lines  are  drawn  to  point  0  in  plane  AB,  making  equal  angles  with 
the  plane.  Find  the  locus  of  point  O.  (HINT.  See  Ex.  1237.) 


470 


SOLID   GEOMETRY 


Ex.  1649.  A  factory  chimney  is  in  the  form  of  a  frustum  of  a  regu- 
lar square  pyramid.  The  chimney  is  120  feet  high,  and  the  edges  of  its 
bases  are  12  feet  and  8  feet,  respectively.  The  flue  is  6  feet  square 
throughout.  How  many  cubic  feet  of  material  does  the  chimney  contain? 

Ex.  1650.  Find  the  edge  of  the  largest  cube  that  can  be  cut  from  a 
regular  square  pyramid  whose  altitude  is  10  inches  and  one  side  of  whose 
base  is  8  inches,  if  one  face  of  the  cube  lies  in  the  base  of  the  pyramid. 

Ex.  1651.  Fig.  1  represents  a  granite  monument,  the  numbers 
denoting  inches.  The  main  part  of  the  stone  is  5  feet  high,  the  total 
height  of  the  stone  being  5  feet  6  inches.  Fig.  2  represents  a  view  of 


54 

FIG.  1. 

the  main  part  of  the  stone  looking  directly  from  above.  Fig.  3  repre- 
sents a  view  of  the  top  of  the  stone  looking  directly  from  above.  Calcu- 
late the  volume  of  the  stone. 

HINT.  From  Fig.  2  it  is  seen  that  the  main  part  of  the  stone  con- 
sists of  a  rectangular  parallelepiped  A,  four  right  triangular  prisms  J5,  and 
a  rectangular  pyramid  at  each  corner.  Fig.  3  shows  that  the  top  con- 
sists of  a  right  triangular  prism  and  two  rectangular  pyramids. 

Ex.  1652.  The  monument  in  Ex.  1651  was  cut  from  a  solid  rock  in 
the  form  of  a  rectangular  parallelepiped.  How  many  cubic  feet  of  granite 
were  wasted  in  the  cutting  ? 

Ex.  1653.  In  the  monument  of  Ex.  1651  the  two  ends  of  the  main 
part,  and  the  top,  have  a  rock  finish,  the  front  and  rear  surfaces  of  the 
main  part  being  polished.  Find  the  number  of  square  feet  of  rock 
finish  and  of  polished  surface. 

Ex.  1654.  The  base  of  a  regular  pyramid  is  a  triangle  inscribed  in  a 
circle  whose  radius  is  J?,  and  the  altitude  of  the  pyramid  is  211.  Find  the 
lateral  area  of  the  pyramid. 

Ex.  1655.  Find  the  weight  in  pounds  of  the  water  required  to  fill  the 
tank  in  Ex.  1323,  if  a  cubic  foot  of  water  weighs  1000  ounces. 


BOOK   IX 


471 


Ex.  1656.  By  using  the  formula  obtained  in  Ex.  1543,  find  the  vol- 
ume of  the  sphere  inscribed  in  a  regular  tetrahedron  whose  edge  is  12. 

Ex.  1657.  By  using  the  formula  obtained  in  Ex.  1544,  find  the  volume 
of  the  sphere  circumscribed  about  a  regular  tetrahedron  whose  edge  is  12. 

Ex.  1658.  A  hole  6  inches  in  diameter  was  bored 
through  a  sphere  10  inches  in  diameter.  Find  the  vol- 
ume of  the  part  cut  out. 

HINT.  The  part  cut  out  consists  of  two  spherical 
cones  and  the  solid  generated  by  revolving  isosceles  A 
BOG  about  XY  as  an  axis. 

Ex.  1659.  Check  your  result  for  Ex.  1658  by 
finding  the  volume  of  the  part  left. 

Ex.   1660.     Find  the  area  of  the  spherical  surface  left  in  Ex.  1658. 

Ex.  1661.  Four  spheres,  each  with  a  radius  of  6  inches,  are  placed 
on  a  plane  surface  in  a  triangular  pile,  each  one  being  tangent  to  each  of 
the  others.  Find  the  total  height  of  the  triangular  pile. 

Ex.  1662.  Find  the  total  height  of  a  triangular  pile  of  spheres,  each 
with  radius  of  6  inches,  if  there  are  three  layers;  four  layers;  n  layers. 


FORMULAS   OF   SOLID   GEOMETRY 

1014.   In  addition  to  the  notation  given  in  §  761,  the  follow- 
ing will  be  used : 


A,  -B,  (7,  •  •  •  =  number  of  degrees  in 
the     angles     of    a 
spherical  polygon. 
a,  6,  c,  •  •  •  =  sides  of    a  spherical 
polygon. 

7?  =  base  of  spherical  sec- 
tor, wedge,  and 
pyramid. 

C  =  circumference  of  base 
in  general  or  of 
lower  base  of  frus- 
tum of  cone. 

c  =  circumference  of  up- 
per base  of  frustum 
of  cone. 

Z>  =  diameter  of  a  sphere. 

E  =  spherical  excess  of  a 
spherical  triangle. 


H  =  altitude  of  zone  or  spherical 

sector. 
K  =  area  of  a  spherical  triangle  or 

spherical  polygon. 
L  =  area  of  lune. 
N  =  number  of  degrees  in  the  angle 

of  a  lune  or  wedge. 
E  =  radius  of  base  in  general,  of 

lower    base   of    frustum   of 

cone,  or  of  sphere. 
r  =  radius  of  upper  base  of  frustum 

of  cone. 

S  =  area  of  surface  of  a  sphere. 
T  =  sum  of  the  angles  of  a  spheri- 
cal polygon. 

W  —  volume  of  a  wedge. 
Z  =  area  of  a  zone. 


472  SOLID   GEOMETRY 

FIGURE  FORMULA  REFERENCE 

Prism.  S  =  P  •  E.  §762. 

Right  prism.  S=P-  H.  §763. 

Regular  pyramid.  8  =  \  P-  L.  §  766. 

Frustum  of  regular  pyramid.  S  =  I  (P .+  p}L.  §  767. 

Rectangular  parallelepiped.  V=  a  •  b  -  c'.  §  778. 

Cube.  V-E^.    '  §779. 


Rectangular  parallelepipeds. 
Rectangular  parallelepiped. 
Rectangular  parallelepipeds. 
Any  parallelepiped. 
Parallelepipeds. 

Triangular  prism. 
Any  prism. 

Prisms. 

Triangular  pyramid. 
Any  pyramid. 

Pyramids. 

Similar  tetrahedrons. 

Frustum  of  any  pyramid. 
Truncated  right  triangular  prism. 
Right  circular  cylinder. 

V       a-b-c 

§780. 
§782. 
§783. 
§790. 
§792. 

§797. 

§799. 

§801. 

§804. 

§805. 

§807. 
§812. 

V     a'-b'  •  c' 
V=B-  H; 
V        B    H 

V      B'  -  H' 
V=B-H. 

v  _  B.H     . 

V      B<  .  H1 
V=B-H. 
V=B-H. 
V        B-H 

V      B'  •  H> 

V=  i  B-  H. 
V        B-H 

V      B>  •  H' 

V  _  E* 
V      E's' 

V=\B(E+E>  +  E" 
8=C-H. 

b).       §815. 
).          §817. 
§858. 
§859. 
§859. 

Similar  cylinders  of  revolution.          —  =  ±L-  =  A. .  §  864. 

/S"      H1'2      R'z 

T^W^W* 

Right  circular  cone.  8  =  \  C  •  L.  §  873. 

S  =  *  R  -  L.  §  875. 

#)•  §  875. 


Similar  cones  of  revolution.  — ,  =  -^  =  ^  =  -^ .  §  878. 


BOOK   IX 


473 


FIGURE 

FORMULA                       REFERENCE 

Similar  cones  of  revolution. 

T   _  H*  _  U-  _  R* 

§878. 

T1      H1'2      L1'2      R1'2 

Frustum  of  right  circular  cone. 

8  =  l(C+c)L. 

§882. 

S  =  irL  (R  +  r}. 

§883. 

- 

T  —  ?rZ,(.R+r)-t-7r(JR2  +  r2) 

.     §  883. 

,     Cylinder  with  circular  bases. 

V=B-H. 

§889. 

V  =  irR2  •  H. 

§890. 

Similar  cylinders  of  revolution. 

Vl  =  Hrs  =  W^ 

§891. 

Cone  with  circular  base. 

V  =  I  B  •  H. 

§893. 

V  =  \  irR*  •  H. 

§895. 

Similar  cones  of  revolution. 

V       Hs       L8       Rs 

§897. 

V  ~  H's  ~~  L'3  ~  R's 

Frustum  of  cone  with  circular  base. 

V  =  1  H(B  +  b  +  V1T&). 

§898. 

V  =  \  irH  (R2  +  r2  •+  R  •  r)  . 

§899. 

Spherical  triangle. 

a  -f  b  >  c. 

§941. 

Spherical  polygon. 

a  +  6  +  c  +  •  •  •  <  360°. 

§942. 

Polar  triangles.                              A 

-f  a'  =  180°,  ^+&'  =  180°,  ••. 

§947. 

Spherical  triangle.                         A 

+  B+  C  >  180°  and  <  540°. 

§949. 

Sphere. 

8  =  4  TrR2. 

§971. 

Spheres. 

S  /_  R2  _  D2 
S'      R1'2     D'2  ' 

§972. 

Zone. 

Z  =  H-2irR. 

§979. 

Zones. 

Z  _H 

§980. 

Z'      H'' 

Lune. 

L  _  N 

§987. 

13     360 

L  =  2N. 

§988. 

Spherical  triangle. 

K  =  (A+B+C)—18QP=E. 

§990. 

Spherical  polygon. 

K=  T—  (n  -2)180. 

§993. 

Sphere. 

V=  S-%R.~ 

§997. 

R  QQQ 

F7?3           713 

§  »yo. 

Spheres. 

-ft           U 
~Vl~  JJ/3  ~~  J)ls' 

§999. 

Spherical  sector. 

V=Z-$R. 

§  1004. 

V  =  |  irR2  •  H. 

§  1005. 

Spherical  wedge. 

W  _  N 

§  1008. 

V      360 

W=L-$R. 

§1009. 

Spherical  triangular  pyramid. 

V=  K-  \R. 

§  1012. 

Any  spherical  pyramid. 

V  —  K-  \R. 

§  1013. 

474 


SOLID   GEOMETRY 


APPENDIX   TO   SOLID   GEOMETRY 
SPHERICAL   SEGMENTS 

1015.  Defs.  A  spherical  segment  is  a  solid  closed  figure 
whose  bounding  surface  consists  of  a  zone  and  two  parallel 
planes. 


Spherical  Segment  of  Two  Bases      Spherical  Segment  of  One  Base 

The  sections  of  the  sphere  formed  by  the  two  parallel  planes 
are  called  the  bases  of  the  spherical  segment. 

1016.  Defs.     State,  by  aid  of  §§  976  and  977,  definitions  of: 
(a)  Altitude  of  a  spherical  segment,      (b)  Segment  of  one  base. 

PROPOSITION  I.     PROBLEM 

1017.  To  derive  a  formula  for  the  volume  of  a  spher- 
ical segment  in  terms  of  the  radii  of  its  bases  and 
its  altitude.  E 


Given  spherical  segment  generated  by  ABCD  revolving  about 
EF  as  an  axis,  with  its  volume  denoted  by  r,  its  altitude  by  h, 
and  the  radii  of  its  bases  by  TI  and  rz,  respectively. 

To  derive   a  formula  for  V  in  terms  of  r1}  r2,  and  h. 


APPENDIX  475 

Draw  radii  OC  and  OD.  Then  F=  volume  of  spherical  sector 
generated  by  COD  +  volume  of  cone  generated  by  BOC  —  vol- 
ume of  cone  generated  by  AOD. 

Denote  OA  by  k,  and  the  radius  of  the  sphere  by  R. 
.-.    v  =  I  irR-li  +  ^  TTI\-  (h  -|-  k)  —  I  TT  r.:k 


But  R2  =  ?y  +  &2  ;  and  R2  =  rf  +  (h  +  fc)2. 

Solving  these  two  equations  for  .R2  and  k, 

2  rffi  -  2  ^  V  r 


1018.    Cor.   I.     Problem.     To  derive  a  formula  for  the 
volume  of  a  spherical  segment  of  one  base  : 

(a)  In  terms  of  its  altitude  and  the  radius  of  its  base  ; 
(6)  In  terms  of  its  altitude  and  the  radius  of  the  sphere. 

(a)  In  §  1017,  put  ^  =  0;  then  F=£uTJJ2/i  +  -J-ir/i3. 

(b)  If  h  represents  the  altitude  of  a  segment  of  one  base, 
and  ?*2  the  radius  of  the  base,  then  r22  =  h  (2  #  —  //,).        §  443,  I. 

Q.E.F. 


Ex.  1663.  A  dumb-bell  consists  of  the  major  portion  of  a  sphere 
with  diameter  6  inches  attached  to  each  end  of  a  right  circular  cylinder 
12  inches  long  and  2  inches  in  diameter.  Find  the  volume  of  the  segment 
cut  from  each  sphere  in  fitting  it  to  the  cylinder. 

Ex.  1664.  By  means  of  the  formulas  given  in  §§  1017  and  1018,  solve 
Exs.  1004  and  1658. 

THE   PRISMATOID 

1019.  Def.     A  prismatoid  is  a  polyhedron  having  for  bases 
two  polygons  in  parallel  planes,  and  for  lateral  faces  triangles 
or  trapezoids  with  one  side  lying  in  one  base,  and  the  opposite 
vertex  or  side  lying  in  the  other  base,  of  the  polyhedron. 

1020.  Def.     The  altitude  of  a  prismatoid  is  the  length  of 
the  perpendicular  from  any  point  in  the  plane  of  one  base  to 
the  plane  of  the  other  base. 


476 


SOLID   GEOMETRY 


PROPOSITION  II.     PROBLEM 
1021.   To    derive    a    formula    for  the    volume    of  a 


prismatoid. 


FIG.  1. 


Given  prismatoid  CF  with  its  volume  denoted  by  v,  its  lower 
base  by  £,  its  upper  base  by  b,  its  altitude  by  H,  and  a  section 
midway  between  the  bases  by  M. 

To  derive   a  formula  for  V  in  terms  of  B,  b,  H,  and  M. 

If  any  lateral  face  as  AD  is  a  trapezoid,  divide  it  into  two  A 
by  diagonal  AD,  intersecting  NK  at  L. 

Let  P  be  any  point  in  M  and  join  it  to  all  vertices  of  the 
prismatoid.  This  will  divide  the  prismatoid  into  pyramids 
having  their  vertices  at  P  and  having  for  their  bases  B,  b,  and 
the  triangles  forming  the  lateral  faces  of  the  prismatoid. 

The  volume  of  pyramid  P-B  —  ^B-^H=^H-B;  and  the 
volume  of  pyramid  P-b  =  %b>\  H  =  ±  H-b.  §  805. 

Consider  pyramid  P-ADC.  Draw  PK,  PL,  and  LC  (Fig.  2). 
This  divides  pyramid  P-ADC  into  three  pyramids,  D-KLP, 
C-KLP,  and  P-ALC.  Denote  A  KLP  by  Wj. 

Then  volume  of  pyramid  D-KLP  =  ^  H  •  mx ;  and  the  vol- 
ume of  pyramid  C-KLP  =  £  J5T  •  w^.  §  805. 


Pyramid  P-ALC 


A  ALC 


but 


A  ALC 


Pyramid  P-CLK  (i.e.  C-KLP)      A  CLK*          A  CLK 
.'.  pyramid  P-ALCo  twice  pyramid  C-KLP. 
.*.  volume  of  pyramid  P-ALC  =  %  H •  m2. 


AC 
LK 


APPENDIX  477 

.  •.  pyramid  P—ADC  =  -J  H  •  m^-J-  H  •  m^  +  177  •  ml  =  ^  H  •  4  Wj. 
.*.  the  volume  of  all  lateral  pyramids  =  -J-  #•  4  3f. 

lf.     Q.E.F. 


Ex.  1665.  By  substituting  in  the  prismatoid  formula,  derive  the 
formula  for  :  (a)  the  volume  of  a  prism  (§  799)  ;  (6)  the  volume  of  a 
pyramid  (§  805)  ;  (c)  the  volume  of  a  frustum  of  a  pyramid  (§  815). 

Ex.  1666.  Solve  Ex.  1651  by  applying  the  prismatoid  formula  to 
each  part  of  the  monument. 

SIMILAR  POLYHEDRONS* 

1022.  The  student  should  prove  the  following  : 

(a)  Any  two  homologous  edges  of  two  similar  polyhedrons  have 
the  same  ratio  as  any  other  two  homologous  edges. 

(b)  Any  two  homologous  faces  of  two  similar  polyhedrons  have 
the  same  ratio  as  the  squares  of  any  two  homologous  edges. 

(c)  The   total   surfaces  of  two   similar  polyhedrons  have  the 
same  ratio  as  the  squares  of  any  two  homologous  edges. 

1023.  Def.     The  ratio  of  similitude  of  two  similar  polyhe- 
drons is  the  ratio  of  any  two  homologous  edges. 

1024.  Def.     If   two  polyhedrons    ABCD  •••  and   A'B'C'D'  ••• 
are  so  situated  that  lines  from  a  point  O  to  A',  B',  C',  D'}  etc., 
are  divided  by  points 

A,  B,  C,  D,  etc.,  in  such 
a  mariner  that 


£4_'  _  OB' 

OA  ~~  OB        OC        OD  E' 

the  two  polyhedrons  are  said  to  be  radially  placed. 


Ex.  1667.     Construct  two  polyhedrons  radially   placed  and  so  that 
point  0  lies  between  the  two  polyhedrons  ;  withm  the  two  polyhedrons. 

*  See  §  811.     In  this  discussion  only  convex  polyhedrons  will  be  considered. 


478 


SOLID   GEOMETRY 


PROPOSITION  III.     THEOREM 

1025.  Any  two  radially  placed  polyhedrons  are 
similar.  (See  Fig.  2  below.) 

Given  polyhedrons  EC  and  E'c'  radially  placed  with  respect 
to  point  0. 

To  prove   polyhedron  EC  ~  polyhedron  E'c'. 

AB,  BC,  CD,  and  DA  are  II  respectively  to  A'ti',  B'c',  C'D',  and 
D'A'.  §  415. 

.-.  ABCD  II  A'B'C'D',  and  is  similar  to  it.  §  756,  II. 

Likewise  each  face  of  polyhedron  EC  is  ~  to  the  correspond- 
ing face  of  polyhedron  E'c',  and  the  faces  are  similarly  placed. 

Again,  face  All  II  face  A'n',  and  face  AF II  face  A'F1. 

.:  dihedral  Z.  AE=  dihedral  Z  A'E1. 

Likewise  each  dihedral  Z  of  polyhedron  EC  is  equal  to  its 
corresponding  dihedral  Z.  of  polyhedron  L'C'. 

.-.  each  polyhedral  Z.  of  polyhedron  EC  is  equal  to  its  corre- 
sponding polyhedral  Z  of  polyhedron  E'c1.  §  18. 

.».  polyhedron  EC  ~  polyhedron  E'C'.     §  811.  Q.E.D. 


PROPOSITION  IY.     THEOREM 

1026.     Any  two  similar  polyhedrons  may  be  radially 
placed. 


N 


Fia.  1. 


FIG.  2. 


Given   two  similar  polyhedrons  XM  and  E'C'. 

To  prove  that  XM  and  E'c'  may  be  radially  placed. 


APPENDIX  479 

OUTLINE  OF  PROOF 

1.  Take  any  point  0  within  polyhedron  E'C1  and  construct 
polyhedron  EC  so  that  it  is  radially  placed  with  respect  to  E'c' 
and  so  that  OA'  :  OA  =  OB1  :  OS  =  •••  =  A'B'  :  KL. 

2.  Then  polyhedron  EC  ~  polyhedron  E'C'.  §  1025. 

3.  Prove  that  the  dihedral  A  of  polyhedron  EC  are  equal, 
respectively,  to  the  dihedral  A  of  polyhedron  XM,  each  being 
equal,  respectively,  to  the  dihedral  A  of  polyhedron  E'c1. 

4.  Prove  that  the  faces  of  polyhedron  EC  are  equal,  respec- 
tively, to  the  faces  of  polyhedron  XM. 

5.  Prove,  by  superposition,  that  polyhedron  EC  =  XM. 

6.  .-.  polyhedron  JfJ/may  be  placed  in  the  position  of  EC. 

7.  But  EC  and  E'C'  are  radially  placed. 

8.  .-.  X M  and  E'c'  may  be  radially  placed.  Q.E.D. 

PROPOSITION  V.     THEOREM 

1027.   If  a  pyramid  is  cut  by  a  plane  parallel  to  its 

base : 

I.  The  pyramid  cut  off  is  similar  to  the  given  pyra- 
mid. 

II.  The  two  pyramids  are  to  each  other  as  the  cubes 
of  any  two  homologous  edges. 

O 


A 

The  proofs  are  left  as  exercises  for  the  student. 

HINT.  For  the  proof  of  II,  pass  planes  through  OS'  and  diagonals 
B'D',  B'E',  etc.,  dividing  each  of  the  pyramids  into  triangular  pyramids. 
Then  pyramid  0-BCD  ~  pyramid  O-B'C'D' ;  pyramid  0-EBD  ~  pyra- 
mid 0-E'B'D',  etc.  Use  §  812  and  a  method  similar  to  that  used  in  §  505. 


480 


SOLID  GEOMETRY 


PROPOSITION  VI.     THEOREM. 

1028.   Two  similar  polyhedrons  are  to  each  other  as 
the  cubes  of  any  two  homologous  edges, 

»' 


Given  two  similar  poly- 
hedrons XM  and  E'c',  with 
their  volumes  denoted  by  V 
and  Ff,  respectively,  and  with  KL  and  A'B'  two  homol.  edges. 

To  prove  —  = 

V' 


Place  XM  in  position  EC,  so  that  XM  and  E'C'  are  radially 
placed  with  respect  to  point  0  within  both  polyhedrons.  §  1026. 

Denote  the  volumes  of  pyramids  0-ABCD,  0-AEFB,  etc.,  by 
VH  vs,  etc.,  and  the  volumes  of  pyramids  0-A'B'c'D1,  0-A'E'F'B1, 
etc.,  by  Vi,  v2',  etc. 


Then 


;  ^=^L;   etc.     §1027,11. 

tt     I  .  I  |O 


AB 


A'B'3      IV       A' 
AE 


AE* 


But      ^  =  -^-=...    (§1022,  a);     .-.  -    -  =  ^-  =  .». 

tint  A '~ci*  ^  '  '  ~~»      io  r      »3 

^1  /j        ^1 JL  A  B         A  E 

V-i   -f-  Va  ~\~  '"  AB  n     t(\-t 

.     l.       V =— -^.    §401. 


AB 


polyhedron  g(7  _  ~Alf  m    .       V__ 
polyhedron  E'c'  ~~  AJB'3  '          F' 


KL 


Q.E.D. 


1029.  Note.  Since  §  1028  was  assumed  early  in  the  text  (see  §814), 
the  teacher  will  find  plenty  of  exercises  throughout  Books  VII,  VIII,  and 
JX  illustrating  this  principle. 


INDEX 


(The  numbers  refer  to  articles.) 


ART. 

Adjacent  dihedral  angles    .  671 

Altitude,  of  cone  .     .     .     .  842 

of  cylinder       ....  825 

of  prism 733 

of  prismatoid  ....  1020 

of  pyramid       ....  751 

of  spherical  segment      .  1016 

of  zone 976 

Angle,  dihedral     ....  666 

magnitude  of  ....  669 

of  lune 983 

'    of  spherical  wedge     .     .  1006 

of  two  intersecting  arcs  916 

polyhedral 692 

solid 692 

spherical 917 

tetrahedral       ....  698 

trihedral 698 

Angles,  designation  of      667,  696 

Axis,  of  circle  of  sphere       .  906 

of  right  circular  cone     .  844 

Base,  of  cone 840 

of  pyramid       ....  748 

of  spherical  pyramid      .  1010 

of  spherical  sector     .     .  1002 

Bases,  of  cylinder  ....  822 

of  prism 727 

of  spherical  segment      .  1015 

of  zone 975 

Birectangular  spherical  tri- 
angle       952 


AKT 

Center  of  sphere   ....  901 

Circular  cone 841 

Circular  cylinder  ....  831 

Circumscribed  polyhedron  .  926 

Circumscribed  prism      .     .  852 

Circumscribed  pyramid       .  868 

Circumscribed  sphere     .     .  929 

Closed  figure    714,  715,  934,  935 

Cone 839 

altitude  of 842 

base  of 840 

circular 841. 

element  of 840 

frustum  of 879 

lateral  surface  of       .     .  840 

oblique 843 

of  revolution    ....  876 

plane  tangent  to  ...  866 

right  circular   ....  843 

spherical 1003 

vertex  of 840 

volume  of   ....     832,  6. 

Cones,  similar 877 

Conical  surface     ....  837 

directrix  of       ....  838 

element  of 838 

generatrix  of    ....  838 

lower  nappe  of      ...  838 

upper  nappe  of     ...  838 

vertex  of 838 

Convex  polyhedral  angle    .  697 

Coplanar  points,  lines,  planes  668 


481 


482 


INDEX 


AET. 

Cube  ........  742 

Cylinder 821 

altitude  of 825 

bases  of 822 

circular 831 

element  of  .....  822 

lateral  surface  of       .     .  822 

oblique 824 

of  revolution    ....  861 

plane  tangent  to  ...  850 

right 823 

right  circular   ....  832 

right  section  of     ...  830 

volume  of 888 

Cylinders,  similar      .     .     .  863 

Cylindrical  surface    .     .     .  819 

directrix  of       ....  820 

element  of 820 

generatrix  of    ....  820 

Degree,  spherical       .     .     .  986 

Determined  plane      .     .     .  608 

Diagonal  of  polyhedron      .  718 

Diameter  of  sphere    .     .     .  901 

Dihedral  angle      ....  666 

edge  of 666 

faces  of 666 

plane  angle  of       ...  670 

right 672 

Dihedral  angles,  adjacent  .  671 

Directrix,  of  conical  surface  838 

of  cylindrical  surface     .  820 

of  polyhedral  angle  .     .  693 

of  prismatic  surface       .  725 

of  pyramidal  surface      .  745 

Distance,  from  point  to  plane  662 

on  surface  of  sphere       .  909 

polar 911 

Dodecahedron       ....  719 

Edge  of  dihedral  angle  .     .  666 

Edges «of  polyhedral  angle  694 


ART. 

Edges  of  polyhedron       .     .  717 

Element,  of  cone  ....  840 

of  conical  surface      .      .  838 

of  cylinder 822 

of  cylindrical  surface     .  820 

of  polyhedral  angle  .     .  694 

of  pyramidal  surface      .  745 

Equivalent  solids       .     .     .  776 

Excess,  spherical  ....  989 

Face  angles  of  polyhedral 

angle 694 

Faces,  of  dihedral  angle       .  666 

of  polyhedral  angle  .     .  694 

of  polyhedron       .     .     .  717 

Foot  of  perpendicular    .     .  621 

Formulas  of  Solid  Geome- 
try       1014 

Frustum  of  cone    ....  879 

slant  height  of      ...  881 

Frustum  of  pyramid       .     .  754 

lateral  area  of       ...  760 

slant  height  of      ...  765 

Generatrix,  of  conical  sur- 
face      838 

of  cylindrical  surface     .  820 

of  polyhedral  angle  .     .  693 

of  prismatic  surface       .  725 

of  pyramidal  surface      .  745 

Geometry,  of  space    .     .     .  602 

solid 602 

Great  circle 904 

Hexahedron 719 

Historical  Notes 

Ahmes 777 

Archimedes      .... 

809,  896,  973,  974,  1001 

Archytas 787 

Athenians   .                   .  787 


INDEX 


483 


ART. 

Historical  Notes 

Brahmagupta  .     .      899,  896 

Cavalieri 992 

Egyptians 777 

Euclid 723 

Eudoxus      .      809,  896,  1000 
Girard,  Albert      .       946,  992 

Hippasus 723 

Menelaus  of  Alexandria  992 

Plato 787 

Pythagoras      .     .       723,  787 

Snell       946 

Socrates 787 

Icosahedron 719 

Inclination  of  line  to  plane  665 

Inscribed  polyhedron      .     .  928 

Inscribed  prism     ....  851 

Inscribed  pyramid     .     .     .  887 

Inscribed  sphere   ....  927 

Intersection  of  two  surfaces  614 

Lateral  area,  of  frustum  of 

pyramid     ....  760 

of  prism 760 

of  pyramid       ....  760 

of  right  circular  cone     .  872 

of  right  circular  cylinder  857 

Lateral  edges,  of  prism  .     .  727 

of  pyramid       ....  748 

Lateral  faces,  of  prism  .     .  727 

of  pyramid       ....  748 

Lateral  surface,  of  cone .     .  840 

of  cylinder 822 

of  frustum  of  pyramid  .  760 

Line,  inclination  of    ...  665 

oblique  to  plane  .     .     .  630 

parallel  to  plane  .     .     .  629 

perpendicular  to  plane  619 

projection  of    ....  656 

tangent  to  sphere      .     .  921 


ART. 

Lune 982 

angle  of       .....  983 

sides  of 983 

vertices  of 983 

Measure-number  ....     770 

Nappes,  upper  and  lower 

746,  838 
Numerical  measure   .  770 


Oblique  cone    . 
Oblique  cylinder 
Octahedron 


843 

824 
719 


Parallel  planes       ....  631 

Parallelepiped       ....  739 

rectangular      .     .     .     .  741 

right 740 

Perpendicular  .     .   619,  620,  672 

foot  of 621 

Perpendicular  planes      .     .  672 

Plane,  determined     .     .     .  608 
perpendicular  to  straight 

line 620 

tangent  to  cone    .     .     .  866 

tangent  to  cylinder  .     .  850 

tangent  to  sphere      .     .  921 
Plane     angle    of     dihedral 

angle 670 

Planes,  parallel     ....  631 

perpendicular  ....  672 

postulate  of      .     .     .     .  615 

Polar  distance  of  circle  .     .  911 

Polar  triangle 943 

Poles  of  circle 907 

Polygon,  spherical     .     .     .  936 

angles  of 936 

diagonal  of       ....  937 

sides  of 936 

vertices  of  .  .  936 


484 


INDEX 


AKT. 

Polyhedral  angle  ....  692 

convex 697 

dihedral  angles  of     .     .  694 

edges  of 694 

element  of  .     .     .     .     .  694 

face  angles  of  ....  694 

faces  of 694 

parts  of 695 

vertex  of      .....  693 

Polyhedral     angles,     sym- 
metrical     ....  707 
vertical 708 

Polyhedron 716 

circumscribed         about 

sphere 926 

diagonal  of       .     .     .     .  718 

edges  of 717 

faces  of  .     .     .     .g    .     .  717 

inscribed  in  sphere    .     .  928 

regular 720 

vertices  of 717 

Polyhedrons,  radially  placed  1024 

similar    .     .     .     .811,  1022 

Portrait  of  Plato  ....  787 

Postulate,  of  planes  .     .     .  615 

revolution 606 

Prism 726 

altitude  of 733 

bases  of 727 

circumscribed  about  cyl- 
inder        852 

inscribed  in  cylinder      .  851 

lateral  area  of       ...  760 

lateral  edges  of     ...  727 

lateral  faces  of      ...  727 

oblique 731 

quadrangular  ....  732 

regular 730 

right 729 

right  section  of     ...  728 

triangular 732 


ABT. 

Prism,  truncated  ....     736 

Prismatic  surface       .     .     .  724 

Prismatoid 1019 

altitude  of 1020 

Projection,  of  line  ....  656 

of  point 655 

Pyramid 747 

altitude  of 751 

base  of 748 

circumscribed  about  cone   868 

frustum  of        ....  754 

inscribed  in  cone       .     .  867 

lateral  area  of       ...  760 

lateral  edges  of     ...  748 

lateral  faces  of      .     .     .  748 

quadrangular  ....  749 

regujar 752 

slant  height  of .     .     .     .  764 

spherical 1010 

triangular    ....  749,  750 

truncated 753 

vertex  of     .....  748 

Pyramidal  surface     .     .     .  744 

Quadrangular  prism       .     .  732 

Quadrangular  pyramid  .     .  749 

Radially  placed  polyhedrons  1024 

Radius  of  sphere  ....  901 

Ratio,  of  similitude    .     .     .  1023 

of  two  solids    ....  775 

Rectangular  parallelepiped  741 

Regular  polyhedron  .     .     .  720 

Regular  prism       ....  730 

Regular  pyramid  ....  752 

Right  circular  cone    .     .     .  843 

axis  of 844 

lateral  area  of       ...  872 

slant  height  of      ...  865 

Right  circular  cylinder  .     .  832 

lateral  area  of  857 


INDEX 


485 


ART. 

Right  cylinder       ....  823 

Right  parallelepiped       .     .  740 

Right  prism 729 

Right  section,  of  cylinder  .  830 

of  prism 728 

Sector,  spherical   ....  1002 
Segment  of  sphere  ....  1015 
Similar  cones  of  revolution    877 
Similar  cylinders  of  revolu- 
tion   863 

Similar  polyhedrons  .  811,  1022 
Similitude,  ratio  of  ...  1023 
Slant  height,  of  frustum  of 

cone 881 

of   frustum  of  pyramid     765 

of  pyramid       ....     764 

of  right  circular  cone     .     865 

Small  circle  of  sphere     .     .     905 

Solid  angle 692 

Solid  geometry      ....     602 

Sphere 900 

center  of 901 

circumscribed        about 

polyhedron  .  .  .  929 
diameter  of  ....  901 
great  circle  of  ...  904 
inscribed  in  polyhedron  927 
line  tangent  to  ...  921 
plane  tangent  to  .  .  .  921 
radius  of  .....  901 
small  circle  of  ...  905 

surface  of 970 

volume  of    .     .     .     .   .  996,  c. 

Spheres,  tangent  externally  .  922 

tangent  internally     .     .     922 

tangent  to  each  other    .     922 

Spherical  angle      .     .     .     .     917 

Spherical  cone       ....  1003 

Spherical  degree    ....     986 

Spherical  excess    ....    989 


AKT. 

Spherical  polygon      .     .     .    936 

angles  of 936 

diagonal  of       ....     937 

sides  of 936 

vertices  of  .     .     .     .     .     936 

Spherical    polygons,    sym- 
metrical     ....     956 

Spherical  pyramid     .     .     .  1010 

base  of 10K 

vertex  of 1010 

Spherical  sector  ....  1002 
base  of 1002 

Spherical  segment      .     .     .1015 

altitude  of 1016 

bases  of 1015 

of  one  base       ....  1016 

Spherical  surface  ....     970 

Spherical  triangle  .  .  .  938 
birectangular  ....  952 
trirectangular  .  .  .  953 

Spherical  wedge    ....  1006 

angle  of       ......  1006 

base  of 1006 

Straight    line,    oblique     to 

plane 630 

parallel  to  plane  .     .     .     629 
perpendicular  to  plane  .     619 

Supplemental  triangles  .     .     948 

Surface,  closed      ....     714 

conical 837 

cylindrical 819 

of  sphere 970 

prismatic 724 

pyramidal 744 

Symmetrical        polyhedral 

angles    .     .     .     .     .     707 

Symmetrical  spherical  poly- 
gons        956 


Tetrahedral  angle 
Tetrahedron 


.  719,  750 


486 


INDEX 


ART. 

Triangle,  polar      ....  943 

spherical 938 

Triangles,  supplemental      .  948 

Triangular  prism 62 

Triangular  pyramid  .     .  749,  750 

Trihedral  angle     ....  698 

birectangular  ....  699 

isosceles 700 

rectangular      ....  699 

trirectangular       .     .     .  699 
Trirectangular        spherical 

triangle       ....  953 

Truncated  prism  ....  736 

Truncated  pyramid  .     .     .  753 

Unit  of  volume     ....  769 

Vertex,  of  cone      ....  840 

of  conical  surface      .     .  838 

of  polyhedral  angle  .     .  693 

of  pyramid       ....  748 


ART. 

Vertex,  of  pyramidal  surface  745 
of  spherical  pyramid  .  1010 

Vertical  polyhedral  angles  .     708 

Vertices,  of  lune  ....  983 
of  polyhedron  .  .  .  717 
of  spherical  polygon  .  936 

Volume,  of  cone    .     .     .     892,  b. 

of  cylinder 888 

of  rectangular  parallele- 
piped       774 

of  solid  .....  769,  770 
of  sphere  ....  996,  c. 
unit  of  .  769 


Wedge,  spherical  . 


1006 


Zone 975 

altitude  of  .     .     .     .     .  976 

bases  ol 975 

of  one  base  977 


ROBBINS'S     PLANE 
TRIGONOMETRY 

By   EDWARD   R.    ROBBINS,  S-iior  Mathematical  Mas- 
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THIS  book  is  intended  for  beginners.      It  aims  to  give  a 
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^[  The  .references  to  Plane  Geometry  preceding  the  first 
chapter  are  invaluable.  A  knowledge  of  the  principles  of 
geometry  needed  in  trigonometry  is,  as  a  rule,  too  freely  taken 
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MILNE'S       STANDARD 
ALGEBRA 

By  WILLIAM  J.   MILNE,  Ph.D.,  LL.D.,   President  of 
the  New  York  State  Normal  College,  Albany,  N.  Y. 

Jl.OO 


THE  Standard  Algebra  conforms  to  the  most  recent 
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except  some  important  ones,  are  left  for  the  maturer  years  of 
the  pupil. 

^[  Accuracy  and  self-reliance  are  encouraged  by  the  use  of 
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ESSENTIALS  OF  BIOLOGY 

By  GEORGE  WILLIAM   HUNTER,  A.  M.,    Head  of 
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IN   this    Manual   the    56   important  problems   of  Hunter's 
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